#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T &a, const T &b) {
  return (a > b ? a = b, 1 : 0);
}
template <class T>
bool umax(T &a, const T &b) {
  return (a < b ? a = b, 1 : 0);
}
void INPUT(bool F = 0) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (F) {
    freopen(
        "name"
        ".in",
        "r", stdin);
    freopen(
        "name"
        ".out",
        "w", stdout);
  }
}
const long long M = 1e9 + 7;
void add(long long &a, long long b) {
  a += b;
  if (a >= M)
    a -= M;
  else if (a < 0)
    a += M;
}
long long mult(long long a, long long b) { return 1ll * a * b % M; }
long long binpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = mult(ans, a);
    b /= 2;
    a = mult(a, a);
  }
  return ans;
}
signed main() {
  INPUT();
  long long n, m, k, r, c;
  cin >> n >> m >> k >> r >> c;
  long long x1, y1, x2, y2;
  cin >> x1 >> y1 >> x2 >> y2;
  long long _x = max(x1, x2), _x1 = min(x1 + r - 1, x2 + r - 1);
  long long _y = max(y1, y2), _y1 = min(y1 + c - 1, y2 + c - 1);
  long long inter =
      max((long long)0, _x1 - _x + 1) * max((long long)0, _y1 - _y + 1);
  long long ans = 1;
  long long free = n * m - 2 * r * c + inter;
  ans = binpow(k, free);
  if (inter != r * c)
    ans = mult(ans, binpow(k, r * c - inter));
  else
    ans = mult(ans, binpow(k, r * c));
  cout << ans;
  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;
template <class T>
bool umin(T &a, const T &b) {
  return (a > b ? a = b, 1 : 0);
}
template <class T>
bool umax(T &a, const T &b) {
  return (a < b ? a = b, 1 : 0);
}
void INPUT(bool F = 0) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (F) {
    freopen(
        "name"
        ".in",
        "r", stdin);
    freopen(
        "name"
        ".out",
        "w", stdout);
  }
}
const long long M = 1e9 + 7;
void add(long long &a, long long b) {
  a += b;
  if (a >= M)
    a -= M;
  else if (a < 0)
    a += M;
}
long long mult(long long a, long long b) { return 1ll * a * b % M; }
long long binpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = mult(ans, a);
    b /= 2;
    a = mult(a, a);
  }
  return ans;
}
signed main() {
  INPUT();
  long long n, m, k, r, c;
  cin >> n >> m >> k >> r >> c;
  long long x1, y1, x2, y2;
  cin >> x1 >> y1 >> x2 >> y2;
  long long _x = max(x1, x2), _x1 = min(x1 + r - 1, x2 + r - 1);
  long long _y = max(y1, y2), _y1 = min(y1 + c - 1, y2 + c - 1);
  long long inter =
      max((long long)0, _x1 - _x + 1) * max((long long)0, _y1 - _y + 1);
  long long ans = 1;
  long long free = n * m - 2 * r * c + inter;
  ans = binpow(k, free);
  if (inter != r * c)
    ans = mult(ans, binpow(k, r * c - inter));
  else
    ans = mult(ans, binpow(k, r * c));
  cout << ans;
  return 0;
}

```