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 mod = 1e9 + 7;
struct Matrix {
  long long c[4][4];
};
long long A, B, n, x;
Matrix Mul(Matrix a, Matrix b) {
  Matrix ans;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      ans.c[i][j] = 0;
      for (int k = 0; k < 2; k++) {
        ans.c[i][j] += a.c[i][k] * b.c[k][j];
        ans.c[i][j] %= mod;
      }
    }
  }
  return ans;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &A, &B, &n, &x);
  Matrix res, tmp;
  memset(res.c, 0, sizeof(res.c));
  memset(tmp.c, 0, sizeof(tmp.c));
  tmp.c[0][0] = A;
  tmp.c[0][1] = B;
  tmp.c[1][1] = 1;
  res.c[0][0] = x;
  res.c[1][0] = 1;
  while (n) {
    if (n & 1) {
      res = Mul(tmp, res);
    }
    tmp = Mul(tmp, tmp);
    n >>= 1;
  }
  printf("%I64d\n", res.c[0][0]);
}
