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;
  void per() { cerr << endl; }
  template <typename Head, typename... Tail>
  void per(Head H, Tail... T) {
    cerr << H << ' ';
    per(T...);
  }
  template <class T>
  bool uin(T& a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T& a, T b) {
    return a < b ? (a = b, true) : false;
  }
  template <class U, class V>
  ostream& operator<<(ostream& out, const pair<U, V>& a) {
    return out << "(" << a.first << ", " << a.second << ")";
  }
  template <class U, class V>
  istream& operator>>(istream& in, pair<U, V>& a) {
    return in >> a.first >> a.second;
  }
  template <typename W,
            typename T = typename enable_if<!is_same<W, string>::value,
                                            typename W::value_type>::type>
  ostream& operator<<(ostream& out, const W& v) {
    out << "{ ";
    for (const auto& first : v) out << first << ", ";
    return out << '}';
  }
  template <class T>
  void readArr(T from, T to) {
    for (auto i = from; i != to; ++i) cin >> *i;
  }
  mt19937 mrand(1337);
  unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
  unsigned long long myRand64() {
    return ((unsigned long long)myRand32() << 32) ^ myRand32();
  }
  const int mod = 1000000007;
  void add(int& a, int b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  void dec(int& a, int b) {
    a -= b;
    if (a < 0) a += mod;
  }
  int mult(int a, int b) { return a * (long long)b % mod; }
  int bp(int a, long long b) {
    int res = 1;
    while (b > 0) {
      if (b & 1) res = mult(res, a);
      a = mult(a, a);
      b >>= 1;
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long n, m, k, r, c;
    cin >> n >> m >> k >> r >> c;
    int ax, ay, bx, by;
    cin >> ax >> ay >> bx >> by;
    if (ax == bx && ay == by) {
      cout << bp(k, n * m) << "\n";
    } else {
      cout << bp(k, n * m - r * c) << "\n";
    }
  }
  