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
import java.util.Scanner;

public class IteratedLinearFunction {

    public static long rightToLeftBinaryExponentiation(long base, long exponent, long modulo) {
        long result = 1;

        while (exponent > 0) {
            if (exponent % 2 == 1) {
                result = (result * base) % modulo;
            }
            base = (base * base) % modulo;
            exponent /= 2;
        }

        return result;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long A = sc.nextLong();
        long B = sc.nextLong();
        long n = sc.nextLong();
        long x = sc.nextLong();
        long modulo = 1000 * 1000 * 1000 + 7;

        if (A == 1) {
            System.out.println((x + (B % modulo) * (n % modulo)) % modulo);
            return;
        }

        long Apowern = rightToLeftBinaryExponentiation(A, n, modulo);
        long AminusOne = rightToLeftBinaryExponentiation(A - 1, modulo - 2, modulo);

        long firstPart = ((Apowern % modulo) * (x % modulo)) % modulo;
        long secondPart = ((B % modulo)
                * ( (((Apowern - 1) % modulo) * (AminusOne % modulo)) % modulo)) % modulo;

        System.out.println((firstPart + secondPart) % modulo);
    }
}
