#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;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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>

### Response

```cpp
#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;
}

```