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 m[2][2];
} ans, bs;
long long m, a, b, x;
matrix multi(matrix a, matrix b) {
  matrix c;
  for (long long i = 0; i < 2; i++) {
    for (long long j = 0; j < 2; j++) {
      c.m[i][j] = 0;
      for (long long k = 0; k < 2; k++) {
        c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j] % mod) % mod;
      }
    }
  }
  return c;
}
long long fast_mod(long long n) {
  bs.m[0][0] = a;
  bs.m[1][1] = bs.m[0][1] = 1;
  ans.m[0][0] = ans.m[1][1] = 1;
  bs.m[1][0] = ans.m[0][1] = ans.m[1][0] = 0;
  while (n) {
    if (n & 1) ans = multi(ans, bs);
    bs = multi(bs, bs);
    n >>= 1;
  }
  return (ans.m[0][0] * x % mod + ans.m[0][1] * b % mod) % mod;
}
int main() {
  while (~scanf("%I64d%I64d%I64d%I64d", &a, &b, &m, &x)) {
    printf("%I64d\n", fast_mod(m));
  }
  return 0;
}
