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
//package assasinskag;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//import java.util.StringTokenizer;
import java.math.*;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;

/* Name of the class has to be "Main" only if the class is public. */
public class feb2
{
	public static void main (String[] args)
	{
		// your code goes here
		Scanner in =new Scanner (System.in);
		int a,b,x;
		long n;
		a=in.nextInt();
		b=in.nextInt();
		
		n=in.nextLong();
		x=in.nextInt();
		if(a==1)
		{
		    BigInteger ans=BigInteger.valueOf(b);
		    ans=ans.multiply(BigInteger.valueOf(n));
		    ans=ans.add(BigInteger.valueOf(x));
		    ans=ans.mod(BigInteger.valueOf(1000000007));
		    	System.out.println(ans);
		    
		}
		else
		{
		BigInteger an=BigInteger.valueOf(a).modPow(BigInteger.valueOf(n),BigInteger.valueOf(1000000007));
		BigInteger aminu=BigInteger.valueOf(a-1).modPow(BigInteger.valueOf(1000000005),BigInteger.valueOf(1000000007));
		BigInteger anx=an.multiply(BigInteger.valueOf(x));
			BigInteger anb=an.multiply(BigInteger.valueOf(b));
			anb=anb.subtract(BigInteger.valueOf(b));
			anb=anb.multiply(aminu);
			anb=anb.add(anx);
			anb=anb.mod(BigInteger.valueOf(1000000007));
			System.out.println(anb);
		}
	}
}
