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 int mod = 1e9 + 7;
long long int power(long long int b, long long int p) {
  if (p == 0) return 1;
  if (p % 2) return (power(b, p - 1) * b) % mod;
  long long int r = power(b, p / 2);
  return (r * r) % mod;
}
long long int inverse(long long int x) { return power(x, mod - 2); }
long long int dof(long long int x) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
int main() {
  long long int a, b, n, x;
  scanf("%lld%lld%lld%lld", &a, &b, &n, &x);
  long long int baki = (power(a, n) * x) % mod, ans;
  if (a == 1) {
    n %= mod, b %= mod;
    ans = (n * b) % mod;
  } else {
    a %= mod, b %= mod;
    ans = (b * dof(power(a, n) - 1)) % mod;
    ans *= inverse(a - 1);
    ans %= mod;
  }
  printf("%lld\n", (ans + baki) % mod);
  return 0;
}
