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;
unsigned long long bigmod(unsigned long long a, unsigned long long b) {
  if (b == 0) return 1;
  unsigned long long ans = bigmod(a, b / 2);
  if (b % 2 == 0) return (ans * ans) % 1000000007;
  return (((ans * ans) % 1000000007) * (a)) % 1000000007;
}
int main() {
  unsigned long long A, B, n, x;
  cin >> A >> B >> n >> x;
  unsigned long long ans = 0;
  unsigned long long bal;
  unsigned long long prothom = bigmod(A, n % (1000000007 - 1));
  bal = prothom;
  prothom = (prothom * x) % 1000000007;
  ans += prothom;
  ans %= 1000000007;
  unsigned long long second;
  if (A != 1)
    second = (bigmod(A - 1, 1000000007 - 2) * (bal - 1)) % 1000000007;
  else
    second = (((n % 1000000007) % 1000000007) * A) % 1000000007;
  second = (second * B) % 1000000007;
  ans = (ans + second) % 1000000007;
  ans %= 1000000007;
  cout << ans;
}
