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.*;
import java.io.*;

public class TestClass{
	
	static PrintWriter out = new PrintWriter(System.out);
	public static void main(String args[]) throws Exception
	{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String s[] = in.readLine().split(" ");
		long a = Long.parseLong(s[0]);
		long b = Long.parseLong(s[1]);
		long n = Long.parseLong(s[2]);
		long x = Long.parseLong(s[3]);
		long mod = (long)1e9+7;
		if(a==1)
		{
			out.println((x+((b%mod)*(n%mod))%mod)%mod);
			out.close();
			System.exit(0);
		}
		long ans = power(a,n,mod);
		ans = (ans*x) % mod;
		long q = (b * (power(a,n,mod)-1) % mod) % mod;
		q = (q * power(a-1,mod-2,mod)%mod)%mod;
		out.println((ans+q)%mod);
		out.close();
	}
	public static long power(long a,long b,long c)
	{
		long ans=1;
		while(b!=0)
		{
			if(b%2==1)
			{
				ans=ans*a;
				ans=ans%c;
			}
			a=a*a;
			a=a%c;
			b=b/2;
		}
		return ans;
	}
}