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 unsigned int pown(long long unsigned int a,
                            long long unsigned int n) {
  long long unsigned int ret = 1;
  a = a % 1000000007;
  while (n > 0) {
    if (n % 2 == 1) ret = (ret * a) % 1000000007;
    n = n >> 1;
    a = (a * a) % 1000000007;
  }
  return ret;
}
long long unsigned int inv(long long unsigned int x) {
  return pown(x, 1000000007 - 2);
}
int main() {
  long long unsigned int a, b, n, x, gn, sec, an;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    cout << (x + (n % 1000000007) * (b % 1000000007)) % 1000000007 << endl;
  } else {
    an = pown(a, n);
    gn = ((an % 1000000007) * (x % 1000000007)) % 1000000007;
    sec = (((an - 1) % 1000000007) * (inv(a - 1) % 1000000007)) % 1000000007;
    sec = ((sec % 1000000007) * (b % 1000000007)) % 1000000007;
    cout << (sec + gn) % 1000000007 << endl;
  }
  return 0;
}
