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.*;
import java.io.*;
import java.math.BigInteger;

public class Tests{
	static Scanner in = new Scanner (System.in);
	//static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	static long MOD = 1000000007;

	static PrintWriter out = new PrintWriter (System.out);

	static long A = in.nextLong(); static long B = in.nextLong();
	static long n = in.nextLong(); static long X = in.nextLong();

	public static void main (String []args) throws IOException,ArithmeticException {
	
		long mat [][] = {{A ,B},{0,1}};
		long ans [][] = pow(mat,n);
		
		out.println(((ans[0][0]*X)%MOD + ans[0][1])%MOD);
		
		

		
		
	   out.flush();
		out.close();
				
	}
	/*
    static long[][] pow(long[][] a, long pow) {
      //  long[][] r = new long[a.length][a.length];
    	long r [][] = {{1,0},{1,0}};
       
        while (pow > 0) {
            if (pow % 2 == 1) {
                r = mul(r, a);
            }
            pow /= 2;
            a = mul(a, a);
        }
        return r;
    }
    */

    static long[][] pow(long a[][], long n) {
        if (n == 1) {
            return a;
        } else if (n % 2 == 0) {
            long t[][] = pow(a, n / 2);
            return mul(t, t);
        } else {
            long t[][] = pow(a, n - 1);
            return mul(a, t);
        }
    }
 
    static long[][] mul (long[][] a, long[][] b) {
        long[][] c = new long[a.length][a.length];
        for (int i = 0; i < a.length; ++i) {
            for (int j = 0; j < a.length; ++j) {
                for (int k = 0; k < a.length; ++k) {
                    c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % MOD;
                }
            }
        }
        return c;
    }
	

	
}