description: Mr. Chanek wants to knit a batik, a traditional cloth from Indonesia. The cloth forms a grid a with size n × m. There are k colors, and each cell in the grid can be one of the k colors.
  
  Define a sub-rectangle as an ordered pair of two cells ((x_1, y_1), (x_2, y_2)), denoting the top-left cell and bottom-right cell (inclusively) of a sub-rectangle in a. Two sub-rectangles ((x_1, y_1), (x_2, y_2)) and ((x_3, y_3), (x_4, y_4)) have the same pattern if and only if the following holds: 
  
    * they have the same width (x_2 - x_1 = x_4 - x_3); 
    * they have the same height (y_2 - y_1 = y_4 - y_3); 
    * for every pair (i, j) where 0 ≤ i ≤ x_2 - x_1 and 0 ≤ j ≤ y_2 - y_1, the color of cells (x_1 + i, y_1 + j) and (x_3 + i, y_3 + j) are equal. 
  
  
  
  Count the number of possible batik color combinations, such that the subrectangles ((a_x, a_y),(a_x + r - 1, a_y + c - 1)) and ((b_x, b_y),(b_x + r - 1, b_y + c - 1)) have the same pattern.
  
  Output the answer modulo 10^9 + 7.
  
  Input
  
  The first line contains five integers n, m, k, r, and c (1 ≤ n, m ≤ 10^9, 1 ≤ k ≤ 10^9, 1 ≤ r ≤ min(10^6, n), 1 ≤ c ≤ min(10^6, m)) — the size of the batik, the number of colors, and size of the sub-rectangle.
  
  The second line contains four integers a_x, a_y, b_x, and b_y (1 ≤ a_x, b_x ≤ n, 1 ≤ a_y, b_y ≤ m) — the top-left corners of the first and second sub-rectangle. Both of the sub-rectangles given are inside the grid (1 ≤ a_x + r - 1, b_x + r - 1 ≤ n, 1 ≤ a_y + c - 1, b_y + c - 1 ≤ m).
  
  Output
  
  Output an integer denoting the number of possible batik color combinations modulo 10^9 + 7.
  
  Examples
  
  Input
  
  
  3 3 2 2 2
  1 1 2 2
  
  
  Output
  
  
  32
  
  
  Input
  
  
  4 5 170845 2 2
  1 4 3 1
  
  
  Output
  
  
  756680455
  
  Note
  
  The following are all 32 possible color combinations in the first example.
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long modn = 1e9 + 7;
  using i64 = long long;
  i64 fastpow(i64 b, i64 p) {
    i64 r = 1LL;
    for (; p > 0; p >>= 1) {
      if (p & 1) r = ((r * b) % modn);
      b = ((b * b) % modn);
    }
    return r;
  }
  long long gcd(long long a, long long b) {
    while (b != 0) {
      long long tmp = a % b;
      a = b;
      b = tmp;
    }
    return a;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    i64 n, m, k, r, c, x1, y1, x2, y2;
    cin >> n >> m >> k >> r >> c >> x1 >> y1 >> x2 >> y2;
    i64 xx = abs(x1 - x2), yy = abs(y1 - y2);
    i64 X = 2 * xx - r, Y = 2 * yy - c;
    if (x1 == x2 && y1 == y2) {
      cout << fastpow(k, n * m);
    } else if (xx >= r && yy >= c) {
      cout << fastpow(k, n * m - r * c);
    } else if (X <= 0 || Y <= 0) {
      cout << fastpow(k, n * m - (2 * r * c - xx * yy) + r * c - xx * yy);
    } else {
      long long s = n * m - (2 * r * c) + (xx * yy);
      s = (s + (r * c - 2 * xx * yy + X * Y));
      s = (s + (xx * yy - X * Y));
      cout << fastpow(k, s);
    }
    return 0;
  }
  