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, x, n;
long long g(long long p, long long x) {
  if (p == 0) return x;
  return (A * g(p - 1, x) % 1000000007 + B % 1000000007) % 1000000007;
}
struct matrix {
  long long c[3][3];
  void init() { memset(c, 0, sizeof(c)); }
  friend matrix operator*(const matrix &S, const matrix &T) {
    matrix res;
    for (int i = (1); i <= (2); i++)
      for (int j = (1); j <= (2); j++) {
        res.c[i][j] = 0;
        for (int k = (1); k <= (2); k++)
          res.c[i][j] = (res.c[i][j] + S.c[i][k] * T.c[k][j]) % 1000000007;
      }
    return res;
  }
} t, ans;
void matrix_pow() {
  ans.c[1][1] = x;
  ans.c[1][2] = 1;
  while (n) {
    if (n & 1) ans = ans * t;
    t = t * t;
    n >>= 1;
  }
}
int main() {
  scanf("%lld%lld%lld%lld", &A, &B, &n, &x);
  t.init();
  ans.init();
  t.c[1][1] = A;
  t.c[2][1] = B;
  t.c[1][2] = 0;
  t.c[2][2] = 1;
  matrix_pow();
  printf("%lld\n", ans.c[1][1]);
  return 0;
}
