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;
int dx[]{1, -1, 0, 0, 1, -1, 1, -1};
int dy[]{0, 0, 1, -1, 1, -1, -1, 1};
const long long mod = 1e9 + 7;
long long pw(long long b, long long p) {
  if (p == 1) return b;
  if (p == 0) return 1;
  long long ret = pw(b, p / 2) % mod;
  ret = (ret * ret) % mod;
  if (p % 2 == 1) ret = (ret * b) % mod;
  return ret;
}
int main() {
  long long n, k, a, b, q, p, x;
  while (cin >> a >> b >> n >> x) {
    if (a == 1)
      cout << ((x % mod) + (b % mod) * (n % mod)) % mod << endl;
    else {
      long long L = (pw(a, n) * x) % mod;
      long long R =
          (((pw(a, n) - 1) % mod) * (pw(a - 1, -1 + mod - 1) % mod)) % mod;
      R = (b * R) % mod;
      cout << (L + R) % mod << endl;
    }
  }
  return 0;
}
