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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modx(long long Base, long long exponent) {
  long long ans = 1;
  if (Base == 1) return Base;
  while (exponent) {
    if (exponent & 1) ans = (ans * Base) % 1000000007;
    Base = (Base * Base) % 1000000007;
    exponent = exponent >> 1;
  }
  return ans;
}
bool cmp() { return false; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, n, x, ans = 0;
  cin >> a >> b >> n >> x;
  if (a == 1)
    ans = (x + ((n % 1000000007) * b) % 1000000007) % 1000000007;
  else
    ans = (((modx(a, n) * x) % 1000000007) +
           ((((b * (modx(a, n) - 1 + 1000000007)) % 1000000007) *
             modx(a - 1 + 1000000007, 1000000007 - 2)) %
            1000000007)) %
          1000000007;
  cout << ans << '\n';
  return 0;
}
