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 IterateLineFunction {
    
    public static final int M = 1000000007;
 
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long A = in.nextLong();
        long B = in.nextLong();
        long n = in.nextLong();
        long x = in.nextLong();
        if (A == 1) {
            System.out.println(modM(x+B*modM(n)));
        } else {
            long An = 1, A2n = A;
            for (int i = 0; i < 63; i++) {
                if (n%2 == 1) {
                    An = modM(An*A2n);
                }
                n >>= 1;
                A2n = modM(A2n*A2n);
            }
            System.out.println(modM(modM(An*x)+modM(B*solvediv(A-1,An-1))));
        }
    }
    
    public static long modM(long a) {
        while(a < 0) {
            a += M;
        }
        return a%M;
    }
    
    public static long solvediv(long a, long b) {
        // ax = b (mod M)
        if (a == 1) {
            return b;
        }
        else {
            return solvediv(M%a,modM(M-b*(M/a)));
        }
    }
    
}