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 pow_mod(long long a, long long i) {
  long long ans = 1;
  while (i) {
    if (i % 2 == 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007;
    i >>= 1;
  }
  return ans;
}
long long getSum(long long a, long long n) {
  return (pow_mod(a, n) - 1) * pow_mod(a - 1, 1000000007 - 2) % 1000000007;
}
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  long long ans = 0;
  if (A == 1) {
    ans = (x + n % 1000000007 * B) % 1000000007;
  } else {
    ans = (pow_mod(A, n) * x % 1000000007 + getSum(A, n) * B % 1000000007) %
          1000000007;
  }
  ans = (ans % 1000000007 + 1000000007) % 1000000007;
  cout << ans << endl;
  return 0;
}
