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.BigInteger;
import java.util.*;

public class iterated_linear_function 
{
	
	public static void main(String args[])throws IOException
	{
		Scanner sc = new Scanner(System.in);
		
		long A = sc.nextLong();
		long B = sc.nextLong();
		long n = sc.nextLong();
		long x = sc.nextLong();
		BigInteger mod = BigInteger.valueOf(1000000000+7);
		
		if (A== 1)
		{
			BigInteger s1 = BigInteger.valueOf(n);
			BigInteger b1 = BigInteger.valueOf(B);
			BigInteger val = (s1.multiply(b1)).add(BigInteger.valueOf(x));
	
			System.out.println(val.mod(mod));
			
		}
		else 
		{
			BigInteger exp = BigInteger.valueOf(n);
			BigInteger Aexp = BigInteger.valueOf(A);
			
			BigInteger A_mod = Aexp.modPow(exp, mod);
			BigInteger s1 = A_mod.multiply(BigInteger.valueOf(x).mod(mod)).mod(mod);
			
			
			BigInteger s2 = BigInteger.valueOf(B).mod(mod);
			BigInteger s3 = A_mod.subtract(BigInteger.valueOf(1).mod(mod));
			BigInteger s4 = (BigInteger.valueOf(A).subtract(BigInteger.valueOf(1))).modInverse(mod);
			BigInteger s5 = ((s2.multiply(s3).multiply(s4)).mod(mod).add(s1)).mod(mod);
			System.out.println(s5);
			
		
		}
	}

}
