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.math.BigInteger;
import java.util.Scanner;

public class IteratedLinearFunction {
	static long pow(long base, long exp)
	{
	    long result = 1;
	    while (exp > 0)
	    {
	        if ((exp & 1) != 0)
	            result = (result*base)%MOD;
	        exp >>= 1;
	        base = ((long)base*base)%MOD;
	    }
	    
	    return result % MOD;
	}
	
	static long inv(long a)
	{
		long res = pow(a, MOD - 2);
		while(res < 0)
			res += MOD;
		
		return res % MOD;
	}
	
	static int MOD = (int) 1e9 + 7;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int A = sc.nextInt();
		int B = sc.nextInt();
		long n = sc.nextLong();
		int x = sc.nextInt();
		
		long res;
		if(A == 1)
			res = (x + (B*(n%MOD))%MOD)%MOD;
		else
			res = ((pow(A,n)*x)%MOD + (B * ((1 - pow(A,n) + MOD)%MOD*inv(1-A)%MOD)%MOD)%MOD)%MOD;
		
		System.out.println(res);
		
		
		
		
	}
}
