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.*;
import java.math.*;

public class Main {
	
	static final BigInteger MOD = new BigInteger("1000000007");
	
	static BigInteger geom(BigInteger a, BigInteger  n){
		BigInteger ans;
		if(n.equals(BigInteger.ZERO)) return BigInteger.ONE;
		if(n.equals(BigInteger.ONE))  return (a.add(BigInteger.ONE)).mod(MOD);
		
		if(a.equals(BigInteger.ONE)){
			return a.multiply(n.add(BigInteger.ONE));
		}
		
		ans = a.modPow(n.add(BigInteger.ONE), MOD);
		ans = ans.subtract(BigInteger.ONE);
		
		BigInteger aux = a.subtract(BigInteger.ONE);
		aux = aux.modInverse(MOD);
		
		ans = ans.multiply(aux);
		
		return ans.mod(MOD);
	}
	
	
	public static void main(String[] args) throws Exception {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		String [] arr = bf.readLine().split(" ");
		BigInteger A,B,x,n;
		
		A = new BigInteger(arr[0]);
		B = new BigInteger(arr[1]);
		n = new BigInteger(arr[2]);
		x = new BigInteger(arr[3]);
		
		BigInteger ans = BigInteger.ONE.multiply(A);
		ans = ans.modPow(n, MOD);
		ans = (ans.multiply(x)).mod(MOD);
		
		BigInteger aux = BigInteger.ONE.multiply(B);
		aux = aux.multiply(geom ( A, n.subtract(BigInteger.ONE) ));
		aux = aux.mod(MOD);
		
		
		System.out.println((ans.add(aux)).mod(MOD));
		
	}
}
