#include <bits/stdc++.h>
using namespace std;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
inline void write(register int x) {
  if (!x) putchar('0');
  if (x < 0) x = -x, putchar('-');
  static int sta[20];
  register int tot = 0;
  while (x) sta[tot++] = x % 10, x /= 10;
  while (tot) putchar(sta[--tot] + 48);
}
int n, m, K, r, c;
int power(int a, long long b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % 1000000007;
    a = 1ll * a * a % 1000000007, b >>= 1;
  }
  return res;
}
int main() {
  n = read(), m = read(), K = read(), r = read(), c = read();
  int i = read(), j = read(), k = read(), l = read();
  if (i == k && j == l)
    printf("%d\n", power(K, 1ll * n * m));
  else
    printf("%d\n", power(K, 1ll * n * m - 1ll * r * c));
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
inline void write(register int x) {
  if (!x) putchar('0');
  if (x < 0) x = -x, putchar('-');
  static int sta[20];
  register int tot = 0;
  while (x) sta[tot++] = x % 10, x /= 10;
  while (tot) putchar(sta[--tot] + 48);
}
int n, m, K, r, c;
int power(int a, long long b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % 1000000007;
    a = 1ll * a * a % 1000000007, b >>= 1;
  }
  return res;
}
int main() {
  n = read(), m = read(), K = read(), r = read(), c = read();
  int i = read(), j = read(), k = read(), l = read();
  if (i == k && j == l)
    printf("%d\n", power(K, 1ll * n * m));
  else
    printf("%d\n", power(K, 1ll * n * m - 1ll * r * c));
  return 0;
}

```