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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void _do(T x) {
  cerr << x << "\n";
}
template <typename T, typename... U>
void _do(T x, U... y) {
  cerr << x << ", ";
  _do(y...);
}
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 3e18;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, k, r, c, x1, x2, y1, y2;
  cin >> n >> m >> k >> r >> c >> x1 >> y1 >> x2 >> y2;
  long long l = max(x1, x2), R = min(x1 + r - 1, x2 + r - 1);
  long long b = max(y1, y2), t = min(y1 + c - 1, y2 + c - 1);
  long long tmp = 0;
  if (l < R && b < t) tmp = (R - l) * (t - b);
  long long sum = r * c * 2 - tmp;
  long long cnt = n * m - sum + r * c - tmp;
  cout << fpow(k, cnt, MOD1);
}
