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;
int64_t phi(int64_t n) {
  int64_t res = n;
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      res = res - res / i;
      do {
        n /= i;
      } while (n % i == 0);
    }
  }
  if (n > 1) res = res - res / n;
  return res;
}
int64_t quickpow_mod(int64_t a, int64_t b, int64_t mod) {
  int64_t res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return res;
}
int main() {
  int64_t a, b, n, x;
  while (cin >> a >> b >> n >> x) {
    int64_t ans = 0;
    int64_t phin = phi(1000000007);
    int64_t res = quickpow_mod(a, n % phin + phin, 1000000007);
    if (a != 1)
      ans = ((res - 1 + 1000000007) % 1000000007 *
                 quickpow_mod(a - 1, 1000000005, 1000000007) % 1000000007 * b %
                 1000000007 +
             (res * x % 1000000007)) %
            1000000007;
    else
      ans = (x + n % 1000000007 * b % 1000000007) % 1000000007;
    cout << ans << endl;
  }
}
