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;
long long powe(long long a, long long b) {
  long long an = 1;
  while (b != 0) {
    if (b % 2 == 1) an = (an * a) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return an;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long an = 0, tmp1, tmpx, tmpy, tmp2;
  tmpx = powe(a, n);
  tmp1 = (tmpx * x) % 1000000007;
  tmpx = (tmpx - 1 + 1000000007) % 1000000007;
  tmp2 = (b * tmpx) % 1000000007;
  tmpy = powe(a - 1, 1000000007 - 2);
  tmp2 = (tmp2 * tmpy) % 1000000007;
  if (a == 1) {
    n = n % 1000000007;
    tmp2 = (n * b) % 1000000007;
  }
  an = (tmp1 + tmp2) % 1000000007;
  cout << an << "\n";
  return 0;
}
