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 int MOD = 1000000007;
  long long modpow(long long x, long long p) {
    if (p == 0) return 1;
    if (p % 2 == 1) return x * modpow(x, p - 1) % MOD;
    long long r = modpow(x, p / 2);
    return r * r % MOD;
  }
  int main() {
    int n, m, k, r, c;
    scanf("%d %d %d %d %d", &n, &m, &k, &r, &c);
    int ax, ay, bx, by;
    scanf("%d %d %d %d", &ax, &ay, &bx, &by);
    if (ax == bx && ay == by) {
      printf("%lld\n", modpow(k, 1LL * n * m));
      return 0;
    }
    long long total = 1LL * n * m;
    long long cnt_union = 2LL * r * c;
    int cx_1 = max(ax, bx);
    int cx_2 = min(ax + r - 1, bx + r - 1);
    int cy_1 = max(ay, by);
    int cy_2 = min(ay + c - 1, by + c - 1);
    if (cx_1 <= cx_2 && cy_1 <= cy_2) {
      long long cnt_intersection = 1LL * (cx_2 - cx_1) * (cy_2 - cy_1);
      cnt_union = cnt_union - cnt_intersection;
    }
    total = total - cnt_union;
    int dx = bx - ax, dy = by - ay;
    long long lx = max(ax, ax + dx);
    long long rx = min(ax + r - 1, ax + r - 1 + dx);
    long long ly = max(ay, ay + dy);
    long long ry = min(ay + c - 1, ay + c - 1 + dy);
    long long add = 1LL * r * c;
    if (lx <= rx && ly <= ry) {
      long long z = (rx - lx) * (ry - ly);
      add = add - z;
    }
    total = total + add;
    printf("%lld\n", modpow(k, total));
    return 0;
  }
  