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 Mat {
  long long a[2][2];
  Mat() { memset(a, 0, sizeof(a)); }
  Mat operator*(const Mat &b) const {
    Mat res;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++) {
          res.a[i][j] += a[i][k] * b.a[k][j];
          res.a[i][j] %= mod;
        }
    return res;
  }
};
Mat bin(Mat &a, long long n) {
  Mat res;
  if (n == 0) {
    res.a[0][0] = res.a[1][1] = 1;
    return res;
  }
  if (n == 1) return a;
  res = bin(a, n >> 1);
  res = res * res;
  if (n & 1) res = res * a;
  return res;
}
int main() {
  Mat a, b;
  long long n;
  scanf("%lld %lld %lld %lld", &a.a[0][0], &a.a[0][1], &n, &b.a[0][0]);
  b.a[1][0] = 1;
  a.a[1][1] = 1;
  a = bin(a, n);
  a = a * b;
  printf("%lld\n", a.a[0][0]);
  return 0;
}
