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.math.BigInteger;
import java.util.Scanner;

/**
 * Created by igarus on 17.10.2016.
 */
public class ECR_13_D {
    public static void main(String[] args) {
        int mod = 1000000007;
        Scanner scanner = new Scanner(System.in);
        int A = scanner.nextInt();
        int B = scanner.nextInt();
        long n = scanner.nextLong();
        int x = scanner.nextInt();


        long an = BigInteger.valueOf(A).modPow(BigInteger.valueOf(n), BigInteger.valueOf(mod)).intValue();
        long c = x * an % mod;

        long r = 0;
        if (A == 1) {
            r = (c + B * (n % mod)) % mod;
        } else {

            long y = B * (an - 1) % mod;
            int z = A - 1;

            long gcdYZ = gcd(y, z);
            y /= gcdYZ;
            z /= gcdYZ;
            if (z == 1) {
                r = (c + y) % mod;
            } else {
                int d = BigInteger.valueOf(z).modPow(BigInteger.valueOf(mod - 2), BigInteger.valueOf(mod)).intValue();
                r = (c + y * d) % mod;
            }
        }

        System.out.println(r);
    }

    public static long gcd(long a, long b) {
        while (b != 0) {
            long tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }
}
