#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
template <class T>
using rque = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  long long cnt = a % b;
  while (cnt != 0) {
    a = b;
    b = cnt;
    cnt = a % b;
  }
  return b;
}
long long extGCD(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
struct UnionFind {
  vector<long long> data;
  int num;
  UnionFind(int sz) {
    data.assign(sz, -1);
    num = sz;
  }
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return (false);
    if (data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    num--;
    return (true);
  }
  int find(int k) {
    if (data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
  long long size(int k) { return (-data[find(k)]); }
  bool same(int x, int y) { return find(x) == find(y); }
};
template <int mod>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt &operator++() { return *this += ModInt(1); }
  ModInt operator++(int) {
    ModInt tmp = *this;
    ++*this;
    return tmp;
  }
  ModInt &operator--() { return *this -= ModInt(1); }
  ModInt operator--(int) {
    ModInt tmp = *this;
    --*this;
    return tmp;
  }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
long long mpow2(long long x, long long n, long long mod) {
  long long ans = 1;
  while (n != 0) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n = n >> 1;
  }
  return ans;
}
long long modinv2(long long a, long long mod) {
  long long b = mod, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= mod;
  if (u < 0) u += mod;
  return u;
}
constexpr int mod = 1000000007;
using mint = ModInt<mod>;
mint mpow(mint x, long long n) {
  mint ans = 1;
  while (n != 0) {
    if (n & 1) ans *= x;
    x *= x;
    n = n >> 1;
  }
  return ans;
}
long long k;
mint solve(long long n, long long m, long long r, long long c) {
  if (n == r && m == c) return mpow(k, n * m);
  if (2 * r - n <= 0 || 2 * c - m <= 0) return mpow(k, n * m - r * c);
  return solve(r, c, 2 * r - n, 2 * c - m) * mpow(k, 2 * (n - r) * (m - c));
}
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(15);
  long long n, m, r, c, ax, ay, bx, by;
  cin >> n >> m >> k >> r >> c >> ax >> ay >> bx >> by;
  if (ax > bx) {
    ax = n + 1 - ax;
    bx = n + 1 - bx;
  }
  if (ay > by) {
    ay = n + 1 - ay;
    by = n + 1 - by;
  }
  long long cn = bx + r - ax, cm = by + c - ay;
  mint ans = mpow(k, n * m - cn * cm);
  ans *= solve(cn, cm, r, c);
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
template <class T>
using rque = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  long long cnt = a % b;
  while (cnt != 0) {
    a = b;
    b = cnt;
    cnt = a % b;
  }
  return b;
}
long long extGCD(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
struct UnionFind {
  vector<long long> data;
  int num;
  UnionFind(int sz) {
    data.assign(sz, -1);
    num = sz;
  }
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return (false);
    if (data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    num--;
    return (true);
  }
  int find(int k) {
    if (data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
  long long size(int k) { return (-data[find(k)]); }
  bool same(int x, int y) { return find(x) == find(y); }
};
template <int mod>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt &operator++() { return *this += ModInt(1); }
  ModInt operator++(int) {
    ModInt tmp = *this;
    ++*this;
    return tmp;
  }
  ModInt &operator--() { return *this -= ModInt(1); }
  ModInt operator--(int) {
    ModInt tmp = *this;
    --*this;
    return tmp;
  }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
long long mpow2(long long x, long long n, long long mod) {
  long long ans = 1;
  while (n != 0) {
    if (n & 1) ans = ans * x % mod;
    x = x * x % mod;
    n = n >> 1;
  }
  return ans;
}
long long modinv2(long long a, long long mod) {
  long long b = mod, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= mod;
  if (u < 0) u += mod;
  return u;
}
constexpr int mod = 1000000007;
using mint = ModInt<mod>;
mint mpow(mint x, long long n) {
  mint ans = 1;
  while (n != 0) {
    if (n & 1) ans *= x;
    x *= x;
    n = n >> 1;
  }
  return ans;
}
long long k;
mint solve(long long n, long long m, long long r, long long c) {
  if (n == r && m == c) return mpow(k, n * m);
  if (2 * r - n <= 0 || 2 * c - m <= 0) return mpow(k, n * m - r * c);
  return solve(r, c, 2 * r - n, 2 * c - m) * mpow(k, 2 * (n - r) * (m - c));
}
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(15);
  long long n, m, r, c, ax, ay, bx, by;
  cin >> n >> m >> k >> r >> c >> ax >> ay >> bx >> by;
  if (ax > bx) {
    ax = n + 1 - ax;
    bx = n + 1 - bx;
  }
  if (ay > by) {
    ay = n + 1 - ay;
    by = n + 1 - by;
  }
  long long cn = bx + r - ax, cm = by + c - ay;
  mint ans = mpow(k, n * m - cn * cm);
  ans *= solve(cn, cm, r, c);
  cout << ans << endl;
}

```