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 mod = 1e9 + 7;
struct Mat {
  long long mat[2][2];
  Mat() { memset(mat, 0, sizeof(mat)); }
};
Mat ec(Mat a, Mat b) {
  Mat c;
  for (int i = 0; i < 2; i++) {
    for (int k = 0; k < 2; k++) {
      for (int j = 0; j < 2; j++) {
        c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j] % mod) % mod;
      }
    }
  }
  return c;
}
Mat ex(Mat a, long long k) {
  Mat b;
  b.mat[0][0] = b.mat[1][1] = 1;
  while (k) {
    if (k % 2 == 1) {
      b = ec(b, a);
    }
    a = ec(a, a);
    k = k / 2;
  }
  return b;
}
int main() {
  long long a, b, x;
  long long n;
  scanf("%I64d%I64d%I64d%I64d", &a, &b, &n, &x);
  Mat c, d, e;
  c.mat[0][0] = a;
  c.mat[0][1] = b;
  c.mat[1][0] = 0;
  c.mat[1][1] = 1;
  d = ex(c, n);
  e.mat[0][0] = x;
  e.mat[1][0] = 1;
  c = ec(d, e);
  printf("%I64d\n", c.mat[0][0]);
  return 0;
}
