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 power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long a, b, x, n;
  cin >> a >> b >> n >> x;
  long long pa = power(a, n);
  long long ans = (pa % 1000000007 * x % 1000000007) % 1000000007;
  long long gpsum = (pa % 1000000007 - 1) % 1000000007;
  if (a == 1)
    gpsum = (n % 1000000007);
  else
    gpsum = (gpsum * modInverse(a % 1000000007 - 1, 1000000007)) % 1000000007;
  ans =
      (ans % 1000000007 + (b % 1000000007 * gpsum % 1000000007) % 1000000007) %
      1000000007;
  cout << ans % 1000000007;
}
