#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long xx, yy, x, y, rr, cc, n, m, k, c, r, xa, ya, xb, yb, p, ans;
long long posw(long long x) {
  long long tt = 1, t = k;
  while (x) {
    if (x & 1) tt = tt * t % MOD;
    x >>= 1;
    t = t * t % MOD;
  }
  return tt;
}
long long sz(long long a, long long b) {
  if (b == 0) return 999999999;
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d%I64d", &n, &m, &k, &r, &c);
  scanf("%I64d%I64d%I64d%I64d", &xa, &ya, &xb, &yb);
  if (xa >= xb && ya >= yb) {
    swap(xa, xb);
    swap(ya, yb);
  }
  if (xa == xb && ya == yb) {
    cout << posw(n * m) << endl;
    return 0;
  }
  if (xb > xa + r - 1 || yb > ya + c - 1) {
    ans = posw(n * m - r * c);
  } else {
    xx = abs(xa - xb);
    yy = abs(ya - yb);
    x = sz(r, xx) - 1;
    y = sz(c, yy) - 1;
    p = min(x, y);
    rr = r - p * xx;
    cc = c - p * yy;
    ans = 2 * (p - 1) * xx * yy - rr * cc +
          (r - (p - 1) * xx) * (c - (p - 1) * yy);
    ans += m * n - (r + xx) * (c + yy) + 2 * xx * yy;
    ans = posw(ans);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 long long MOD = 1e9 + 7;
long long xx, yy, x, y, rr, cc, n, m, k, c, r, xa, ya, xb, yb, p, ans;
long long posw(long long x) {
  long long tt = 1, t = k;
  while (x) {
    if (x & 1) tt = tt * t % MOD;
    x >>= 1;
    t = t * t % MOD;
  }
  return tt;
}
long long sz(long long a, long long b) {
  if (b == 0) return 999999999;
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d%I64d", &n, &m, &k, &r, &c);
  scanf("%I64d%I64d%I64d%I64d", &xa, &ya, &xb, &yb);
  if (xa >= xb && ya >= yb) {
    swap(xa, xb);
    swap(ya, yb);
  }
  if (xa == xb && ya == yb) {
    cout << posw(n * m) << endl;
    return 0;
  }
  if (xb > xa + r - 1 || yb > ya + c - 1) {
    ans = posw(n * m - r * c);
  } else {
    xx = abs(xa - xb);
    yy = abs(ya - yb);
    x = sz(r, xx) - 1;
    y = sz(c, yy) - 1;
    p = min(x, y);
    rr = r - p * xx;
    cc = c - p * yy;
    ans = 2 * (p - 1) * xx * yy - rr * cc +
          (r - (p - 1) * xx) * (c - (p - 1) * yy);
    ans += m * n - (r + xx) * (c + yy) + 2 * xx * yy;
    ans = posw(ans);
  }
  cout << ans << endl;
  return 0;
}

```