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;
unsigned long long MOD = 1000000007;
unsigned long long mod = MOD;
unsigned long long pow(unsigned long long a, unsigned long long b) {
  unsigned long long x = 1LL, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int main() {
  unsigned long long a, b, n, x;
  cin >> a >> b >> n >> x;
  unsigned long long l = ((b % MOD) * (pow(a, n) + MOD - 1) % MOD) % MOD;
  unsigned long long r = (a - 1);
  unsigned long long ans = (pow(a, n) * (x % mod));
  if (r == 0) {
    cout << (ans % MOD + ((b % MOD) * (n % MOD)) % MOD) % MOD << endl;
  } else {
    ans += (((b % mod * (pow(a, n) - 1 + mod) % mod) % mod * pow(r, mod - 2)) %
            mod) %
           mod;
    ans %= MOD;
    cout << ans << endl;
  }
  return 0;
}
