description: 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>
solution: #include <bits/stdc++.h>
  template <class T>
  using MinHeap = std::priority_queue<T, std::vector<T>, std::greater<T>>;
  template <class T>
  using MaxHeap = std::priority_queue<T>;
  using ll = long long;
  using Pii = std::pair<int, int>;
  using Pll = std::pair<ll, ll>;
  using Pdd = std::pair<double, double>;
  template <class T>
  bool chmin(T &a, const T b) {
    if (a > b) {
      a = b;
      return true;
    } else {
      return false;
    }
  }
  template <class T>
  bool chmax(T &a, const T b) {
    if (a < b) {
      a = b;
      return true;
    } else {
      return false;
    }
  }
  template <class T>
  void vdeb(const std::vector<T> &da) {
    auto n = da.size();
    for (size_t i = 0; i < n; i++) {
      if (i == n - 1)
        std::cout << da[i];
      else
        std::cout << da[i] << " ";
    }
    std::cout << std::endl;
  }
  template <class T>
  void vdeb(const std::vector<std::vector<T>> &da) {
    auto n = da.size();
    for (size_t i = 0; i < n; i++) {
      std::cout << i << " : ";
      vdeb(da[i]);
    }
  }
  template <>
  void vdeb(const std::vector<std::string> &da) {
    auto n = da.size();
    for (size_t i = 0; i < n; i++) {
      std::cout << da[i] << std::endl;
    }
  }
  using namespace std;
  const ll MOD = 1000000007;
  ll modpow(ll x, ll k) {
    ll ret = 1, now = x;
    while (k) {
      if (k & 1) {
        ret *= now;
        ret %= MOD;
      }
      now = now * now % MOD;
      k >>= 1;
    }
    return ret;
  }
  int main() {
    ll n, m, k, r, c;
    cin >> n >> m >> k >> r >> c;
    ll ax, ay, bx, by;
    cin >> ax >> ay >> bx >> by;
    cout << ((ax == bx && ay == by) ? modpow(k, n * m) : modpow(k, n * m - r * c))
         << endl;
  }
  