description: The statement of this problem shares a lot with problem A. The differences are that in this problem, the probability is introduced, and the constraint is different.
  
  A robot cleaner is placed on the floor of a rectangle room, surrounded by walls. The floor consists of n rows and m columns. The rows of the floor are numbered from 1 to n from top to bottom, and columns of the floor are numbered from 1 to m from left to right. The cell on the intersection of the r-th row and the c-th column is denoted as (r,c). The initial position of the robot is (r_b, c_b).
  
  In one second, the robot moves by dr rows and dc columns, that is, after one second, the robot moves from the cell (r, c) to (r + dr, c + dc). Initially dr = 1, dc = 1. If there is a vertical wall (the left or the right walls) in the movement direction, dc is reflected before the movement, so the new value of dc is -dc. And if there is a horizontal wall (the upper or lower walls), dr is reflected before the movement, so the new value of dr is -dr.
  
  Each second (including the moment before the robot starts moving), the robot cleans every cell lying in the same row or the same column as its position. There is only one dirty cell at (r_d, c_d). The job of the robot is to clean that dirty cell. 
  
  After a lot of testings in problem A, the robot is now broken. It cleans the floor as described above, but at each second the cleaning operation is performed with probability \frac p {100} only, and not performed with probability 1 - \frac p {100}. The cleaning or not cleaning outcomes are independent each second.
  
  Given the floor size n and m, the robot's initial position (r_b, c_b) and the dirty cell's position (r_d, c_d), find the expected time for the robot to do its job.
  
  It can be shown that the answer can be expressed as an irreducible fraction \frac x y, where x and y are integers and y not ≡ 0 \pmod{10^9 + 7} . Output the integer equal to x ⋅ y^{-1} mod (10^9 + 7). In other words, output such an integer a that 0 ≤ a < 10^9 + 7 and a ⋅ y ≡ x \pmod {10^9 + 7}.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
  
  A test case consists of only one line, containing n, m, r_b, c_b, r_d, c_d, and p (4 ≤ n ⋅ m ≤ 10^5, n, m ≥ 2, 1 ≤ r_b, r_d ≤ n, 1 ≤ c_b, c_d ≤ m, 1 ≤ p ≤ 99) — the sizes of the room, the initial position of the robot, the position of the dirt cell and the probability of cleaning in percentage.
  
  Output
  
  For each test case, print a single integer — the expected time for the robot to clean the dirty cell, modulo 10^9 + 7.
  
  Example
  
  Input
  
  
  6
  2 2 1 1 2 1 25
  3 3 1 2 2 2 25
  10 10 1 1 10 10 75
  10 10 10 10 1 1 75
  5 5 1 3 2 2 10
  97 98 3 5 41 43 50
  
  
  Output
  
  
  3
  3
  15
  15
  332103349
  99224487
  
  Note
  
  In the first test case, the robot has the opportunity to clean the dirty cell every second. Using the [geometric distribution](https://en.wikipedia.org/wiki/Geometric_distribution), we can find out that with the success rate of 25\%, the expected number of tries to clear the dirty cell is \frac 1 {0.25} = 4. But because the first moment the robot has the opportunity to clean the cell is before the robot starts moving, the answer is 3.
  
  <image> Illustration for the first example. The blue arc is the robot. The red star is the target dirt cell. The purple square is the initial position of the robot. Each second the robot has an opportunity to clean a row and a column, denoted by yellow stripes.
  
  In the second test case, the board size and the position are different, but the robot still has the opportunity to clean the dirty cell every second, and it has the same probability of cleaning. Therefore the answer is the same as in the first example.
  
  <image> Illustration for the second example.
  
  The third and the fourth case are almost the same. The only difference is that the position of the dirty cell and the robot are swapped. But the movements in both cases are identical, hence the same result.
solution: import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.*;
  import static java.util.Arrays.*;
  
  public class cf1623d {
  
      public static void main(String[] args) throws IOException {
          int T = ri();
          next: while (T --> 0) {
              int n = rni(), m = ni(), rb = ni(), cb = ni(), rd = ni(), cd = ni(), p = ni(), ip = 100 - p;
              p = mmul(p, minv(100));
              ip = mmul(ip, minv(100));
              int rx, rx2, ry, ry2;
              if (rd < rb) {
                  rx = rb - rd + (n - rb) * 2;
                  rx2 = (n - rb) + (n - 1) + (rd - 1);
              } else {
                  rx = rd - rb;
                  rx2 = rd - rb + (n - rd) * 2;
              }
              if (cd < cb) {
                  ry = cb - cd + (m - cb) * 2;
                  ry2 = (m - cb) + (m - 1) + (cd - 1);
              } else {
                  ry = cd - cb;
                  ry2 = cd - cb + (m - cd) * 2;
              }
              n = 2 * (n - 1);
              m = 2 * (m - 1);
              int cycle = n / gcd(n, m) * m;
              rx %= n;
              rx2 %= n;
              ry %= m;
              ry2 %= m;
              Set<Integer> keyframes = new TreeSet<>();
              for (int i = 0; i < cycle / n; ++i) {
                  keyframes.add(rx + n * i);
                  keyframes.add(rx2 + n * i);
              }
              for (int i = 0; i < cycle / m; ++i) {
                  keyframes.add(ry + m * i);
                  keyframes.add(ry2 + m * i);
              }
              int cnt = keyframes.size(), cur = 0, e0 = 0;
              for (int t : keyframes) {
                  if (t >= cycle) {
                      break;
                  }
                  e0 = madd(e0, mmul(t, mpow(ip, cur++), p));
              }
              // prln(rx, rx2, ry, ry2);
              // prln(keyframes);
              // prln(cycle, cnt, mmul(e0, 256));
              // (e0 + cycle * ip ** cnt) / (1 - ip ** cnt)
              prln(mmul(madd(e0, mmul(cycle, mpow(ip, cnt))), minv(msub(1, mpow(ip, cnt)))));
          }
          close();
      }
  
      static int mmod = 1000000007;
  
      static int madd(int a, int b) {
          return (a + b) % mmod;
      }
  
      static int madd(int... a) {
          int ans = a[0];
          for (int i = 1; i < a.length; ++i) {
              ans = madd(ans, a[i]);
          }
          return ans;
      }
  
      static int msub(int a, int b) {
          return (a - b + mmod) % mmod;
      }
  
      static int mmul(int a, int b) {
          return (int) ((long) a * b % mmod);
      }
  
      static int mmul(int... a) {
          int ans = a[0];
          for (int i = 1; i < a.length; ++i) {
              ans = mmul(ans, a[i]);
          }
          return ans;
      }
  
      static int minv(int x) {
          // return mpow(x, mmod - 2);
          return (exgcd(x, mmod)[0] % mmod + mmod) % mmod;
      }
  
      static int mpow(int a, long b) {
          if (a == 0) {
              return 0;
          }
          int ans = 1;
          while (b > 0) {
              if ((b & 1) > 0) {
                  ans = mmul(ans, a);
              }
              a = mmul(a, a);
              b >>= 1;
          }
          return ans;
      }
  
      static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
      static StringTokenizer input;
      static Random __r = new Random();
  
      // references
      // IBIG = 1e9 + 7
      // IMAX ~= 2e9
      // LMAX ~= 9e18
  
      // constants
      static final int IBIG = 1000000007;
      static final int IMAX = 2147483647;
      static final long LMAX = 9223372036854775807L;
      // math util
      static int minof(int a, int b, int c) {return min(a, min(b, c));}
      static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
      static long minof(long a, long b, long c) {return min(a, min(b, c));}
      static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
      static int maxof(int a, int b, int c) {return max(a, max(b, c));}
      static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
      static long maxof(long a, long b, long c) {return max(a, max(b, c));}
      static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
      static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
      static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
      static int fli(double d) {return (int) d;}
      static int cei(double d) {return (int) ceil(d);}
      static long fll(double d) {return (long) d;}
      static long cel(double d) {return (long) ceil(d);}
      static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
      static long gcd(long a, long b) {return b == 0 ? a : gcd(b, a % b);}
      static int[] exgcd(int a, int b) {if (b == 0) return new int[] {1, 0, a}; int[] y = exgcd(b, a % b); return new int[] {y[1], y[0] - y[1] * (a / b), y[2]};}
      static long[] exgcd(long a, long b) {if (b == 0) return new long[] {1, 0, a}; long[] y = exgcd(b, a % b); return new long[] {y[1], y[0] - y[1] * (a / b), y[2]};}
      static int randInt(int min, int max) {return __r.nextInt(max - min + 1) + min;}
      static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
      // array util
      static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
      static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
      static void rsort(int[] a) {shuffle(a); sort(a);}
      static void rsort(long[] a) {shuffle(a); sort(a);}
      static void rsort(double[] a) {shuffle(a); sort(a);}
      static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
      // input
      static void r() throws IOException {input = new StringTokenizer(rline());}
      static int ri() throws IOException {return Integer.parseInt(rline());}
      static long rl() throws IOException {return Long.parseLong(rline());}
      static double rd() throws IOException {return Double.parseDouble(rline());}
      static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
      static void ria(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni();}
      static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
      static void riam1(int[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1;}
      static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
      static void rla(long[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nl();}
      static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
      static void rda(double[] a) throws IOException {int n = a.length; r(); for (int i = 0; i < n; ++i) a[i] = nd();}
      static char[] rcha() throws IOException {return rline().toCharArray();}
      static void rcha(char[] a) throws IOException {int n = a.length, i = 0; for (char c : rline().toCharArray()) a[i++] = c;}
      static String rline() throws IOException {return __i.readLine();}
      static String n() {return input.nextToken();}
      static int rni() throws IOException {r(); return ni();}
      static int ni() {return Integer.parseInt(n());}
      static long rnl() throws IOException {r(); return nl();}
      static long nl() {return Long.parseLong(n());}
      static double rnd() throws IOException {r(); return nd();}
      static double nd() {return Double.parseDouble(n());}
      // output
      static void pr(int i) {__o.print(i);}
      static void prln(int i) {__o.println(i);}
      static void pr(long l) {__o.print(l);}
      static void prln(long l) {__o.println(l);}
      static void pr(double d) {__o.print(d);}
      static void prln(double d) {__o.println(d);}
      static void pr(char c) {__o.print(c);}
      static void prln(char c) {__o.println(c);}
      static void pr(char[] s) {__o.print(new String(s));}
      static void prln(char[] s) {__o.println(new String(s));}
      static void pr(String s) {__o.print(s);}
      static void prln(String s) {__o.println(s);}
      static void pr(Object o) {__o.print(o);}
      static void prln(Object o) {__o.println(o);}
      static void prln() {__o.println();}
      static void pryes() {prln("yes");}
      static void pry() {prln("Yes");}
      static void prY() {prln("YES");}
      static void prno() {prln("no");}
      static void prn() {prln("No");}
      static void prN() {prln("NO");}
      static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
      static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
      static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
      static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
      static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
      static void h() {prln("hlfd"); flush();}
      static void flush() {__o.flush();}
      static void close() {__o.close();}
  }