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;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author root
 */
public class D {

    static long mod = 1000000007L;

    public static void main(String argsp[]) {
        Scanner in = new Scanner(System.in);
        long A = in.nextLong();
        long B = in.nextLong();
        long n = in.nextLong();
        long x = in.nextLong();
        long An = pow(A, n);
        long REV = pow(A - 1, (mod - 2)) % mod;
        long Part = (((An - 1) % mod) * (REV % mod)) % mod;
        long ans;
        if (A == 1) {
            ans = (((n%mod)*(B%mod))%mod+((An%mod)*(x%mod))%mod)%mod;
        } else {
            ans = ((((An % mod) * (x % mod)) % mod) + ((B % mod) * (Part) % mod) % mod) % mod;
        }
        System.out.println(ans);
    }

    public static long pow(long a, long p) {
        if (a == 1L) {
            return 1L;
        }
        if (p == 1) {
            return a;
        }
        long ans = 0;
        long pow = p / 2;
        long x = a % mod;
        if (p % 2 == 0) {
            return pow(x * x, p / 2);
        }
        return ((x % mod) * (pow(x * x, p / 2) % mod)) % mod;
    }
}
