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;
  using ll = long long;
  constexpr int INF = 0x3f3f3f3f;
  constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
  constexpr double EPS = 1e-8;
  constexpr int MOD = 1000000007;
  constexpr int DY[]{1, 0, -1, 0}, DX[]{0, -1, 0, 1};
  constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1},
      DX8[]{0, -1, -1, -1, 0, 1, 1, 1};
  template <typename T, typename U>
  inline bool chmax(T &a, U b) {
    return a < b ? (a = b, true) : false;
  }
  template <typename T, typename U>
  inline bool chmin(T &a, U b) {
    return a > b ? (a = b, true) : false;
  }
  struct IOSetup {
    IOSetup() {
      std::cin.tie(nullptr);
      std::ios_base::sync_with_stdio(false);
      std::cout << fixed << setprecision(20);
    }
  } iosetup;
  template <int M>
  struct MInt {
    unsigned int val;
    MInt() : val(0) {}
    MInt(long long x) : val(x >= 0 ? x % M : x % M + M) {}
    static constexpr int get_mod() { return M; }
    static void set_mod(int divisor) { assert(divisor == M); }
    static void init(int x = 10000000) {
      inv(x, true);
      fact(x);
      fact_inv(x);
    }
    static MInt inv(int x, bool init = false) {
      static std::vector<MInt> inverse{0, 1};
      int prev = inverse.size();
      if (init && x >= prev) {
        inverse.resize(x + 1);
        for (int i = prev; i <= x; ++i) inverse[i] = -inverse[M % i] * (M / i);
      }
      if (x < inverse.size()) return inverse[x];
      unsigned int a = x, b = M;
      int u = 1, v = 0;
      while (b) {
        unsigned int q = a / b;
        std::swap(a -= q * b, b);
        std::swap(u -= q * v, v);
      }
      return u;
    }
    static MInt fact(int x) {
      static std::vector<MInt> f{1};
      int prev = f.size();
      if (x >= prev) {
        f.resize(x + 1);
        for (int i = prev; i <= x; ++i) f[i] = f[i - 1] * i;
      }
      return f[x];
    }
    static MInt fact_inv(int x) {
      static std::vector<MInt> finv{1};
      int prev = finv.size();
      if (x >= prev) {
        finv.resize(x + 1);
        finv[x] = inv(fact(x).val);
        for (int i = x; i > prev; --i) finv[i - 1] = finv[i] * i;
      }
      return finv[x];
    }
    static MInt nCk(int n, int k) {
      if (n < 0 || n < k || k < 0) return 0;
      if (n - k > k) k = n - k;
      return fact(n) * fact_inv(k) * fact_inv(n - k);
    }
    static MInt nPk(int n, int k) {
      return n < 0 || n < k || k < 0 ? 0 : fact(n) * fact_inv(n - k);
    }
    static MInt nHk(int n, int k) {
      return n < 0 || k < 0 ? 0 : (k == 0 ? 1 : nCk(n + k - 1, k));
    }
    static MInt large_nCk(long long n, int k) {
      if (n < 0 || n < k || k < 0) return 0;
      inv(k, true);
      MInt res = 1;
      for (int i = 1; i <= k; ++i) res *= inv(i) * n--;
      return res;
    }
    MInt pow(long long exponent) const {
      MInt tmp = *this, res = 1;
      while (exponent > 0) {
        if (exponent & 1) res *= tmp;
        tmp *= tmp;
        exponent >>= 1;
      }
      return res;
    }
    MInt &operator+=(const MInt &x) {
      if ((val += x.val) >= M) val -= M;
      return *this;
    }
    MInt &operator-=(const MInt &x) {
      if ((val += M - x.val) >= M) val -= M;
      return *this;
    }
    MInt &operator*=(const MInt &x) {
      val = static_cast<unsigned long long>(val) * x.val % M;
      return *this;
    }
    MInt &operator/=(const MInt &x) { return *this *= inv(x.val); }
    bool operator==(const MInt &x) const { return val == x.val; }
    bool operator!=(const MInt &x) const { return val != x.val; }
    bool operator<(const MInt &x) const { return val < x.val; }
    bool operator<=(const MInt &x) const { return val <= x.val; }
    bool operator>(const MInt &x) const { return val > x.val; }
    bool operator>=(const MInt &x) const { return val >= x.val; }
    MInt &operator++() {
      if (++val == M) val = 0;
      return *this;
    }
    MInt operator++(int) {
      MInt res = *this;
      ++*this;
      return res;
    }
    MInt &operator--() {
      val = (val == 0 ? M : val) - 1;
      return *this;
    }
    MInt operator--(int) {
      MInt res = *this;
      --*this;
      return res;
    }
    MInt operator+() const { return *this; }
    MInt operator-() const { return MInt(val ? M - val : 0); }
    MInt operator+(const MInt &x) const { return MInt(*this) += x; }
    MInt operator-(const MInt &x) const { return MInt(*this) -= x; }
    MInt operator*(const MInt &x) const { return MInt(*this) *= x; }
    MInt operator/(const MInt &x) const { return MInt(*this) /= x; }
    friend std::ostream &operator<<(std::ostream &os, const MInt &x) {
      return os << x.val;
    }
    friend std::istream &operator>>(std::istream &is, MInt &x) {
      long long val;
      is >> val;
      x = MInt(val);
      return is;
    }
  };
  namespace std {
  template <int M>
  MInt<M> abs(const MInt<M> &x) {
    return x;
  }
  }  // namespace std
  using ModInt = MInt<MOD>;
  int main() {
    int n, m, k, r, c, ax, ay, bx, by;
    cin >> n >> m >> k >> r >> c >> ax >> ay >> bx >> by;
    if (ax == bx && ay == by) {
      cout << ModInt(k).pow(1LL * n * m) << '\n';
    } else {
      cout << ModInt(k).pow(1LL * n * m - 1LL * r * c) << '\n';
    }
    return 0;
  }
  