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.*;

public class IteratedLinearFunction
{
	
	static long mod = (long)(1e9+7);

	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		long a = scan.nextLong();
		long b = scan.nextLong();
		long n = scan.nextLong();
		long x = scan.nextLong();
		if(a == 1) {
			long ans = b*(n%mod)%mod;
			ans += x;
			ans %= mod;
			System.out.println(ans);
			return;
		}
		long A = (pow(a, n)*x)%mod;
		long B = (b*(pow(a,n)-1)%mod)%mod;
		long minv = ((a-1)%mod+mod)%mod;
		minv = modInv(minv);
//		System.out.println(minv);
		B = B*minv;
		B %= mod;
//		System.out.println(A+" "+B);
		System.out.println((A+B)%mod);
	}
	
	static long modInv(long b) {
		return pow(b, mod-2);
	}
	
	static long pow(long b, long p) {
		if(p == 0) return 1;
		if(p % 2 == 1) {
			long x = pow(b, p-1);
			return (x*b)%mod;
		}
		long x = pow(b, p/2);
		return (x*x)%mod;
	}
}
