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 M = 1e9 + 7;
struct matrix {
  long long v[2][2];
} A, c;
matrix operator*(const matrix& a, const matrix& b) {
  memset(c.v, 0, sizeof c.v);
  for (int i = 0; i < 2; ++i)
    for (int j = 0; j < 2; ++j)
      for (int k = 0; k < 2; ++k)
        c.v[i][j] = (c.v[i][j] + a.v[i][k] * b.v[k][j]) % M;
  return c;
}
long long x, n;
matrix Pow(matrix a, long long b) {
  matrix s;
  s.v[0][0] = s.v[1][1] = 1;
  s.v[0][1] = s.v[1][0] = 0;
  while (b) {
    if (b & 1) s = s * a;
    b = b / 2;
    a = a * a;
  }
  return s;
}
int main() {
  cin >> A.v[0][0] >> A.v[0][1] >> n >> x;
  A.v[1][1] = 1;
  A = Pow(A, n);
  cout << (A.v[0][0] * x % M + A.v[0][1]) % M;
}
