#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int maxn = 2e6 + 10;
const int mod = 1e9 + 7;
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) { return ksm(a, mod - 2); }
int not_prime[maxn], prime[maxn], tot;
void phii() {
  for (int i = 2; i <= maxn; i++) {
    if (!not_prime[i]) prime[++tot] = i;
    for (int j = 1; j <= tot && prime[j] * i <= maxn; j++) {
      not_prime[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
signed main() {
  long long n, m, k, r, c, A, B, C, D;
  cin >> n >> m >> k >> r >> c;
  cin >> A >> B >> C >> D;
  if (A == C && B == D)
    cout << ksm(k, n * m) << endl;
  else
    cout << ksm(k, n * m - r * c) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 double eps = 1e-8;
const int maxn = 2e6 + 10;
const int mod = 1e9 + 7;
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) { return ksm(a, mod - 2); }
int not_prime[maxn], prime[maxn], tot;
void phii() {
  for (int i = 2; i <= maxn; i++) {
    if (!not_prime[i]) prime[++tot] = i;
    for (int j = 1; j <= tot && prime[j] * i <= maxn; j++) {
      not_prime[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
signed main() {
  long long n, m, k, r, c, A, B, C, D;
  cin >> n >> m >> k >> r >> c;
  cin >> A >> B >> C >> D;
  if (A == C && B == D)
    cout << ksm(k, n * m) << endl;
  else
    cout << ksm(k, n * m - r * c) << endl;
  return 0;
}

```