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 long long mod = 1e9 + 7;
long long quickpow(long long m, long long n, long long k) {
  long long b = 1;
  while (n > 0) {
    if (n & 1) b = (b * m) % k;
    n = n >> 1;
    m = (m * m) % k;
  }
  return b;
}
int main(void) {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    cout << (n % mod * b + x) % mod << endl;
    return 0;
  }
  long long ans = (quickpow(a, n, mod) - 1 + mod) % mod;
  ans = ans * quickpow(a - 1, mod - 2, mod) % mod * b % mod;
  ans = (ans + quickpow(a, n, mod) * x + mod) % mod;
  printf("%lld\n", ans % mod);
}
