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 long long mod = 1000000007LL;
struct matrix {
  long long m[2][2];
  matrix() { m[0][0] = m[0][1] = m[1][0] = m[1][1] = 0; }
  void unit() { m[0][0] = m[1][1] = 1; }
};
matrix mul_mod(matrix a, matrix b) {
  matrix ans;
  int i, j, k;
  for (i = 0; i < 2; ++i)
    for (j = 0; j < 2; ++j) {
      for (k = 0; k < 2; ++k) ans.m[i][j] += a.m[i][k] * b.m[k][j];
      ans.m[i][j] %= mod;
    }
  return ans;
}
matrix pow_mod(matrix a, long long n) {
  matrix ans;
  ans.unit();
  while (n) {
    if (n & 1) ans = mul_mod(ans, a);
    a = mul_mod(a, a);
    n >>= 1;
  }
  return ans;
}
int main() {
  matrix M;
  M.m[1][1] = 1;
  scanf("%I64d%I64d", &M.m[0][0], &M.m[0][1]);
  long long n, x;
  scanf("%I64d%I64d", &n, &x);
  x %= mod;
  M = pow_mod(M, n);
  printf("%I64d\n", (M.m[0][0] * x + M.m[0][1]) % mod);
  return 0;
}
