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 powmod(long long a, long long b) {
  if (b == 0) return 1;
  long long x = powmod(a, b / 2);
  long long y = (x * x) % 1000000007;
  if (b % 2) return (a * y) % 1000000007;
  return y % 1000000007;
}
long long mulmod(long long a, long long b, long long c) {
  long long x = 0, y = a % c;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x + y) % c;
    }
    y = (y * 2) % c;
    b /= 2;
  }
  return x % c;
}
int main() {
  long long a, b, x, n;
  cin >> a >> b >> n >> x;
  long long nm1, nm2, nm3, nm4, nm5;
  if (a > 1) {
    nm1 = powmod(a, n);
    nm2 = (nm1 - 1 + 1000000007) % 1000000007;
    nm3 = (x * nm1) % 1000000007;
    nm4 = (a - 1 + 1000000007) % 1000000007;
    nm5 = (nm2 * powmod(nm4, 1000000007 - 2)) % 1000000007;
    long long nm6 = (nm3 + (nm5 * b) % 1000000007) % 1000000007;
    cout << nm6 << endl;
  } else
    cout << (x + mulmod(b, n, 1000000007)) % 1000000007 << endl;
  return 0;
}
