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 mod = 1000000007;
long long int pwr(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if ((b & 1) == 1) res = ((res % mod) * (a % mod)) % mod;
    b >>= 1;
    a = ((a % mod) * (a % mod)) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long int p1, p2, a, b, x, n, result, an, bn, dn;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    result = ((x % mod) + ((n % mod) * (b % mod)) % mod) % mod;
    cout << result;
    return 0;
  }
  an = pwr(a, n);
  dn = pwr(a - 1, mod - 2);
  bn = ((an - 1) % mod + mod) % mod;
  bn = ((bn % mod) * (b % mod)) % mod;
  result = ((an % mod) * (x % mod)) % mod;
  bn = ((bn % mod) * (dn % mod)) % mod;
  result = ((result % mod) + (bn % mod)) % mod;
  cout << result;
  return 0;
}
