Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers A, B, n and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Examples

Input

3 4 1 1


Output

7


Input

3 4 2 1


Output

25


Input

3 4 3 1


Output

79
#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long phi(long long m) {
  long long r = m;
  for (int i = 2; i * i <= m; i++) {
    if (m % i == 0) {
      while (m % i == 0) m /= i;
      r -= r / i;
    }
  }
  if (m > 1) r -= r / m;
  return r;
}
long long powe(long long a, long long b) {
  long long m = 1;
  while (b) {
    if (b % 2)
      m = (m * a) % M, b--;
    else
      b /= 2, a = (a * a) % M;
  }
  return m;
}
int main() {
  long long a, b, n, x, ans = 0, N, q;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    long long ans = (x + ((b % M) * (n % M)) % M) % M;
    cout << ans;
    return 0;
  }
  N = phi(M) - 1;
  q = ((powe(a, n) - 1) * powe(a - 1, N)) % M;
  ans = (q * b) % M;
  ans += (powe(a, n) * x) % M;
  cout << ans % M;
}
