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>
  using namespace std;
  template <int MOD, int RT>
  struct Mint {
    static const int mod = MOD;
    static constexpr Mint rt() { return RT; }
    static constexpr int md() { return MOD; }
    int v;
    explicit operator int() const { return v; }
    explicit operator bool() const { return v != 0; }
    Mint() { v = 0; }
    Mint(long long _v) {
      v = int((-MOD <= _v && _v < MOD) ? _v : _v % MOD);
      if (v < 0) v += MOD;
    }
    friend bool operator==(const Mint& a, const Mint& b) { return a.v == b.v; }
    friend bool operator!=(const Mint& a, const Mint& b) { return !(a == b); }
    friend bool operator<(const Mint& a, const Mint& b) { return a.v < b.v; }
    friend bool operator>(const Mint& a, const Mint& b) { return a.v > b.v; }
    friend bool operator<=(const Mint& a, const Mint& b) { return a.v <= b.v; }
    friend bool operator>=(const Mint& a, const Mint& b) { return a.v >= b.v; }
    friend std::istream& operator>>(std::istream& in, Mint& a) {
      long long x;
      std::cin >> x;
      a = Mint(x);
      return in;
    }
    friend std::ostream& operator<<(std::ostream& os, const Mint& a) {
      return os << a.v;
    }
    Mint& operator+=(const Mint& m) {
      if ((v += m.v) >= MOD) v -= MOD;
      return *this;
    }
    Mint& operator-=(const Mint& m) {
      if ((v -= m.v) < 0) v += MOD;
      return *this;
    }
    Mint& operator*=(const Mint& m) {
      v = (long long)v * m.v % MOD;
      return *this;
    }
    Mint& operator/=(const Mint& m) { return (*this) *= inv(m); }
    friend Mint pow(Mint a, long long p) {
      Mint ans = 1;
      assert(p >= 0);
      for (; p; p /= 2, a *= a)
        if (p & 1) ans *= a;
      return ans;
    }
    friend Mint inv(const Mint& a) {
      assert(a.v != 0);
      return pow(a, MOD - 2);
    }
    Mint operator-() const { return Mint(-v); }
    Mint& operator++() { return *this += 1; }
    Mint& operator--() { return *this -= 1; }
    friend Mint operator+(Mint a, const Mint& b) { return a += b; }
    friend Mint operator-(Mint a, const Mint& b) { return a -= b; }
    friend Mint operator*(Mint a, const Mint& b) { return a *= b; }
    friend Mint operator/(Mint a, const Mint& b) { return a /= b; }
  };
  using mi = Mint<1000000007, 5>;
  int main() {
    long long n, m, k, r, c;
    cin >> n >> m >> k >> r >> c;
    int w, x, y, z;
    cin >> w >> x >> y >> z;
    mi ans;
    if (w == y && x == z) {
      ans = pow(mi(k), n * m);
    } else {
      ans = pow(mi(k), n * m - r * c);
    }
    cout << ans << '\n';
    return 0;
  }
  