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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;


public class ProblemD {
	static long MOD = 1000000007;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		long A = input.nextLong();
		long B = input.nextLong();
		long N = input.nextLong();
		long X = input.nextLong();
		//value of G0(X) = X
		//modular exponentiation all day long
		ArrayList<Integer> modularExponentiation = new ArrayList<Integer>();
		while(N > 0){
			modularExponentiation.add((int)(N%2));
			N = N / 2;
		}
		long[] arrA = new long[100];
		long[] arrB = new long[100];
		arrA[0] = A;
		arrB[0] = B;
		for(int a = 1; a < 100; a++){
			arrA[a] = arrA[a-1] * arrA[a-1];
			arrB[a] = arrA[a-1] * arrB[a-1] + arrB[a-1];
			arrA[a] %= MOD;
			arrB[a] %= MOD;
		}
		long start = X;
		for(int a = 0; a < modularExponentiation.size(); a++){
			if(modularExponentiation.get(a) == 1){
				start = arrA[a]*start + arrB[a];
				start = start%MOD;
			}
		}
		System.out.println(start);
	}
}
