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 = 1000000007;
struct matrix {
  long long m[2][2];
};
long long A, B, x, n;
matrix mul(matrix a, matrix b) {
  matrix tmp;
  memset(tmp.m, 0, sizeof(tmp.m));
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++)
        tmp.m[i][j] = tmp.m[i][j] + a.m[i][k] * b.m[k][j] % mod;
      tmp.m[i][j] %= mod;
    }
  return tmp;
}
long long cal(long long h) {
  matrix a, b, ans;
  ans.m[0][0] = x;
  ans.m[1][1] = a.m[0][1] = 1;
  ans.m[1][0] = B;
  b.m[0][0] = A;
  b.m[1][0] = 0;
  b.m[0][1] = b.m[1][1] = 1;
  a.m[0][0] = a.m[1][1] = 1;
  a.m[1][0] = a.m[0][1] = 0;
  while (h > 0) {
    if (h & 1) a = mul(a, b);
    b = mul(b, b);
    h >>= 1;
  }
  ans = mul(a, ans);
  return ans.m[0][0];
}
int main() {
  while (~scanf("%I64d%I64d%I64d%I64d", &A, &B, &n, &x)) {
    if (n == 0)
      printf("%I64d\n", x % mod);
    else
      printf("%I64d\n", cal(n) % mod);
  }
  return 0;
}
