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>
  const int MOD = 1e9 + 7;
  int n, m, k, r, c;
  int ax, ay, bx, by;
  long long fpow(long long a, long long b) {
    long long ret = 1;
    while (b) {
      if (b & 1) ret = ret * a % MOD;
      a = a * a % MOD;
      b >>= 1;
    }
    return ret;
  }
  bool inter(int x1, int x2, int x3, int x4) {
    if (x1 > x3) {
      std::swap(x1, x3);
      std::swap(x2, x4);
    }
    return x1 <= x3 && x3 <= x2;
  }
  int main() {
    scanf("%d%d%d%d%d", &n, &m, &k, &r, &c);
    scanf("%d%d%d%d", &ax, &ay, &bx, &by);
    if (ax == bx && ay == by) {
      printf("%lld\n", fpow(k, (long long)n * m));
      return 0;
    }
    if (!inter(ax, ax + r - 1, bx, bx + r - 1) ||
        !inter(ay, ay + c - 1, by, by + c - 1)) {
      long long ans = fpow(k, (long long)n * m - (long long)r * c);
      printf("%lld\n", ans);
      return 0;
    }
    if (ax > bx) {
      std::swap(ax, bx);
      std::swap(ay, by);
    }
    if (ay > by) {
      int diff = by + c - 1 - ay + 1;
      by = ay + c - 1 - diff + 1;
    }
    long long inter_area =
        (long long)(ay + c - 1 - by + 1) * (ax + c - 1 - bx + 1);
    long long graph_area = 2 * (long long)r * c - inter_area;
    long long out_area = (long long)n * m - graph_area;
    long long valid_area = (long long)r * c - inter_area;
    printf("%lld\n", fpow(k, out_area + valid_area));
    return 0;
  }
  