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;
using namespace std;
long long power(long long x, long long n) {
  if (n < 0) return power(x, -n);
  if (n == 0) return 1;
  if (n == 1) return x % 1000000007;
  if (n & 1)
    return ((x % 1000000007) *
            power(((x % 1000000007) * (x % 1000000007)) % 1000000007,
                  (n - 1) / 2)) %
           1000000007;
  return power(((x % 1000000007) * (x % 1000000007)) % 1000000007, n / 2) %
         1000000007;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long aPowerN = power(a, n);
  long long leftPart = ((aPowerN % 1000000007) * (x % 1000000007)) % 1000000007;
  long long num = (aPowerN % 1000000007 - 1 % 1000000007) % 1000000007;
  long long dem = (a % 1000000007 - 1 % 1000000007) % 1000000007;
  long long demInv = power(dem, 1000000007 - 2);
  long long rightPart =
      ((num % 1000000007) * (demInv % 1000000007)) % 1000000007;
  if (a == 1) rightPart = n % 1000000007;
  cout << (leftPart % 1000000007 +
           (rightPart * (b % 1000000007)) % 1000000007) %
              1000000007;
}
