#include <bits/stdc++.h>
using namespace std;
using ld = long double;
using ll = long long;
ll mod = 1e9 + 7;
int const N = 3e5 + 100;
mt19937 rnd;
ll random(ll limit) {
  uniform_int_distribution<ll> uniform_dist(0, limit);
  return uniform_dist(rnd);
}
ll bpow(ll a, ll p) {
  ll res = 1;
  while (p > 0) {
    if (p % 2) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    p /= 2;
  }
  return res;
}
ll inv(ll a) { return bpow(a, mod - 2); }
void solve() {
  ll n, m, k, r, c;
  cin >> n >> m >> k >> r >> c;
  ll x1, x2, y1, y2;
  cin >> x1 >> y1 >> x2 >> y2;
  ll dx = min(abs(x1 - x2), c);
  ll dy = min(abs(y1 - y2), r);
  ll num1 = dx * r + dy * (c - dx);
  if (dx == 0 && dy == 0) {
    num1 = r * c;
  } else if (dx == 0) {
    num1 = c * dy;
  } else if (dy == 0)
    num1 = r * dx;
  ll x3 = max(x1, x2);
  ll x4 = min(x1 + c, x2 + c);
  ll y3 = max(y1, y2);
  ll y4 = min(y1 + r, y2 + r);
  ll num3 = (x4 - x3) * (y4 - y3);
  if (y4 <= y3 || x4 <= x3) num3 = 0;
  ll num2 = n * m - (2 * r * c - num3);
  ll num = num1 + num2;
  ll ans = bpow(k, num);
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  rnd = mt19937(956231);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in cpp 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;
using ld = long double;
using ll = long long;
ll mod = 1e9 + 7;
int const N = 3e5 + 100;
mt19937 rnd;
ll random(ll limit) {
  uniform_int_distribution<ll> uniform_dist(0, limit);
  return uniform_dist(rnd);
}
ll bpow(ll a, ll p) {
  ll res = 1;
  while (p > 0) {
    if (p % 2) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    p /= 2;
  }
  return res;
}
ll inv(ll a) { return bpow(a, mod - 2); }
void solve() {
  ll n, m, k, r, c;
  cin >> n >> m >> k >> r >> c;
  ll x1, x2, y1, y2;
  cin >> x1 >> y1 >> x2 >> y2;
  ll dx = min(abs(x1 - x2), c);
  ll dy = min(abs(y1 - y2), r);
  ll num1 = dx * r + dy * (c - dx);
  if (dx == 0 && dy == 0) {
    num1 = r * c;
  } else if (dx == 0) {
    num1 = c * dy;
  } else if (dy == 0)
    num1 = r * dx;
  ll x3 = max(x1, x2);
  ll x4 = min(x1 + c, x2 + c);
  ll y3 = max(y1, y2);
  ll y4 = min(y1 + r, y2 + r);
  ll num3 = (x4 - x3) * (y4 - y3);
  if (y4 <= y3 || x4 <= x3) num3 = 0;
  ll num2 = n * m - (2 * r * c - num3);
  ll num = num1 + num2;
  ll ans = bpow(k, num);
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  rnd = mt19937(956231);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```