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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.StringTokenizer;


public class Task678D {

    public static void main(String[] args) throws IOException {
        Scanner is = new Scanner(System.in);
        long a_in = is.nextLong();
        long b_in = is.nextLong();
        long n_in = is.nextLong();
        long x_in = is.nextLong();
        long m = 1000000007;
        BigInteger a = new BigInteger(String.valueOf(a_in));
        BigInteger b = new BigInteger(String.valueOf(b_in));
        BigInteger n = new BigInteger(String.valueOf(n_in));
        BigInteger x = new BigInteger(String.valueOf(x_in));
        BigInteger mod = new BigInteger(String.valueOf(m));

        BigInteger p = a.modPow(n, mod).multiply(x);
        if (a_in == 1) p  = p.add(b.multiply(n).mod(mod)).mod(mod);
        else {
            BigInteger a_1 = new BigInteger(String.valueOf(a_in - 1));
            p =  p.add(b.mod(mod).multiply((a.modPow(n, mod)).subtract(BigInteger.ONE)).mod(mod).multiply(a_1.modInverse(mod)));
        }
        System.out.println(p.mod(mod).toString());
    }
}
