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 por(long long a, long long b) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > 1000000007) x %= 1000000007;
    }
    y = (y * y);
    if (y > 1000000007) y %= 1000000007;
    b /= 2;
  }
  return x;
}
int main() {
  long long i, j, k, a, b, n, x, p;
  ios::sync_with_stdio(0);
  cin >> a >> b >> n >> x;
  if (a == 1) {
    j = ((b * ((n) % 1000000007) + x) % 1000000007) % 1000000007;
  } else if (n == 1)
    j = ((a * x) % 1000000007 + b) % 1000000007;
  else {
    j = por(a, n);
    p = j;
    k = j * x;
    p--;
    if (p < 0) p += 1000000007;
    j = por(a - 1, 1000000007 - 2);
    j = ((j * p) % 1000000007 * b) % 1000000007;
    j = (j + k) % 1000000007;
  }
  cout << j << endl;
}
