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;
const long long mod = 1000000007;
array<array<long long, 2>, 2> multiply(array<array<long long, 2>, 2>& left,
                                       array<array<long long, 2>, 2>& right) {
  array<array<long long, 2>, 2> result;
  for (auto i = 0; i != 2; i++) {
    for (auto j = 0; j != 2; j++) {
      auto sum = 0ll;
      for (auto k = 0; k != 2; k++) {
        sum += left[i][k] * right[k][j];
      }
      result[i][j] = sum % mod;
    }
  }
  return result;
}
array<array<long long, 2>, 2> exp(array<array<long long, 2>, 2> M,
                                  long long n) {
  array<array<long long, 2>, 2> result = {{{1, 0}, {0, 1}}};
  while (n != 0) {
    if (n % 2 == 1) {
      result = multiply(result, M);
    }
    M = multiply(M, M);
    n /= 2;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  array<array<long long, 2>, 2> M = {{{A, 1}, {0, 1}}};
  M = exp(M, n);
  cout << (M[0][0] * x + M[0][1] * B) % mod;
  return 0;
}
