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>
#include <bits/stdc++.h>
using namespace std;
long double PI = acos(-1);
long double DEL = 1e-12;
long long M = 1e9 + 7;
const long long N = 3e5 + 10;
long long cnt, sum, mx, mn, ans, a[N], b[N];
long long n, m, d, i, j, k, l, p, q, r, c, t, w, x, y, z, tc;
string s;
long long fpow(long long x, long long n) {
  long long res = 1;
  x %= M;
  while (n) {
    if (n & 1) res = res * x % M;
    x = x * x % M;
    n >>= 1;
  }
  return res;
}
long long noofways(long long n, long long m, long long r, long long c) {
  if (n >= 2 * r or m >= 2 * c) {
    return fpow(k, n * m - r * c);
  }
  long long ext = (n * m) - 2 * r * c + (2 * r - n) * (2 * c - m);
  return noofways(r, c, 2 * r - n, 2 * c - m) * fpow(k, ext) % M;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k >> r >> c;
  long long ax, ay, bx, by;
  cin >> ax >> ay >> bx >> by;
  if (ax == bx and ay == by) {
    cout << fpow(k, n * m);
    return 0;
  }
  if (ax > bx) {
    swap(ax, bx);
  }
  if (ay > by) {
    swap(ay, by);
  }
  long long ext = n * m - (by + c - ay) * (bx + r - ax);
  cout << fpow(k, ext) * noofways(bx + r - ax, by + c - ay, r, c) % M;
}
