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 a, b, n, x, M[2][2];
void multiply(long long C[2][2], long long A[2][2], long long B[2][2]) {
  int i, j, k;
  for (i = 0; i < 2; ++i)
    for (j = 0; j < 2; ++j) {
      C[i][j] = 0;
      for (k = 0; k < 2; ++k)
        C[i][j] += A[i][k] * B[k][j] % 1000000007, C[i][j] %= 1000000007;
    }
}
void mcopy(long long B[2][2], long long A[2][2]) {
  int i, j;
  for (i = 0; i < 2; ++i)
    for (j = 0; j < 2; ++j) B[i][j] = A[i][j];
}
void pow(long long A[2][2], long long n) {
  long long sol[2][2] = {{1, 0}, {0, 1}}, aux[2][2], B[2][2];
  mcopy(B, A);
  while (n) {
    if (n & 1) {
      multiply(aux, sol, B);
      mcopy(sol, aux);
    }
    multiply(aux, B, B);
    mcopy(B, aux);
    n >>= 1;
  }
  mcopy(A, sol);
}
int main() {
  cin >> a >> b >> n >> x;
  M[0][0] = a, M[0][1] = b, M[1][1] = 1;
  pow(M, n);
  cout << (M[0][0] * x % 1000000007 + M[0][1]) % 1000000007 << "\n";
  return 0;
}
