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 N = 100005, mod = 1e9 + 7;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int i, j, k, n, m, ans = 0, cnt = 0, sum = 0, x, a, b;
    cin >> a >> b >> n >> x;
    cnt = power(a, n);
    ans = (cnt * x) % mod;
    if (a == 1)
      ans = (ans + (b * ((n % mod * a) % mod)) % mod) % mod;
    else {
      sum = ((cnt - 1) * power(a - 1, mod - 2)) % mod;
      ans = (ans + (sum * b) % mod) % mod;
    }
    cout << ans;
  }
  return 0;
}
