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 int pw[5111];
long long int fact[5111];
long long int ifact[5111];
long long int bpow(long long int x, long long int n) {
  long long int ans = 1;
  while (n > 0) {
    if (n & 1) ans *= x;
    x *= x;
    ans %= 1000000007;
    x %= 1000000007;
    n /= 2;
  }
  return ans;
}
int main(int argc, char const *argv[]) {
  long long int a, b, n, x;
  cin >> a >> b >> n >> x;
  long long int ans;
  if (a == 1) {
    ans = (x + n % 1000000007 * b % 1000000007) % 1000000007;
  } else {
    long long int pow1 = bpow(a, n);
    long long int pow2 = bpow(a - 1, 1000000007 - 2);
    ans = (pow1 * x + (((pow1 - 1) % 1000000007 * b % 1000000007) % 1000000007 *
                       pow2 % 1000000007) %
                          1000000007) %
          1000000007;
  }
  cout << ans << endl;
  return 0;
}
