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 i64 = long long;
  const int MOD = 1000000007;
  i64 pow64(i64 a, i64 b) {
    i64 result = 1;
    while (b) {
      if (b & 1) result = result * a % MOD;
      a = a * a % MOD;
      b >>= 1;
    }
    return result;
  }
  int n, m, k, r, c;
  int ax, ay, bx, by;
  int f(int a, int b, int c, int d) {
    if (a >= c * 2 || b >= d * 2) return (i64(a) * b - i64(c) * d) % (MOD - 1);
    if (a == c) {
      if (b == d)
        return i64(a) * b % (MOD - 1);
      else {
        int dy = b - d;
        int cc = (d + dy - 1) / dy - 1;
        return f(a, b - cc * dy, c, d - cc * dy);
      }
    } else {
      if (b == d)
        return f(b, a, d, c);
      else {
        int dx = a - c, dy = b - d;
        int cc = std::min((c + dx - 1) / dx, (d + dy - 1) / dy) - 1;
        return (f(a - cc * dx, b - cc * dy, c - cc * dx, d - cc * dy) +
                2 * cc * i64(dx) * dy) %
               (MOD - 1);
      }
    }
  }
  int main() {
    std::scanf("%d%d%d%d%d", &n, &m, &k, &r, &c);
    std::scanf("%d%d%d%d", &ax, &ay, &bx, &by);
    int xl = std::min(ax, bx), xr = std::max(ax, bx) + r;
    int yl = std::min(ay, by), yr = std::max(ay, by) + c;
    int dx = xr - xl, dy = yr - yl;
    int result = (i64(n) * m - i64(dx) * dy + f(dx, dy, r, c)) % (MOD - 1);
    result = pow64(k, result);
    std::printf("%d\n", result);
  }
  