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 mod(long long a, long long b) { return ((a % b) + b) % b; }
long long extended_euclid(long long a, long long b, long long& x,
                          long long& y) {
  long long current_x = y = 0;
  long long current_y = x = 1;
  while (b) {
    long long q = a / b;
    long long t = b;
    b = a % b;
    a = t;
    t = current_x;
    current_x = x - q * current_x;
    x = t;
    t = current_y;
    current_y = y - q * current_y;
    y = t;
  }
  return a;
}
long long mod_inverse(long long a, long long n) {
  long long x, y;
  long long d = extended_euclid(a, n, x, y);
  if (d > 1) return -1;
  return mod(x, n);
}
long long A, B, n, x;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> A >> B >> n >> x;
  if (A == 1) {
    B = ((n % 1000000007) * B) % 1000000007;
    cout << (x + B) % 1000000007 << endl;
  } else {
    long long i = 1;
    long long res = 1;
    long long p = A;
    while (i <= n) {
      if (i & n) {
        res = (res * p) % 1000000007;
      }
      i = i * 2;
      p = (p * p) % 1000000007;
    }
    long long inv = mod_inverse(A - 1, 1000000007);
    long long ans = ((res * x) % 1000000007) +
                    (((B * inv) % 1000000007) * (res - 1)) % 1000000007;
    ans = ((ans % 1000000007) + 1000000007) % 1000000007;
    cout << ans << endl;
  }
}
