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 probD {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		long a = sc.nextLong();
		long b = sc.nextLong();
		long n = sc.nextLong();
		long x = sc.nextLong();

		BigInteger aa = new BigInteger("" + a);
		BigInteger aaa = new BigInteger("" + (a - 1));
		BigInteger bb = new BigInteger("" + b);
		BigInteger nn = new BigInteger("" + n);
		BigInteger xx = new BigInteger("" + x);
		BigInteger mod = new BigInteger("1000000007");
		
		if(a==1)
		{
			BigInteger bbb = xx.add((bb.multiply(nn)).mod(mod));
			bbb = bbb.mod(mod);
			System.out.println(bbb);
			return;
		}

		BigInteger ans = new BigInteger("0");

		ans = aa.modPow(nn, mod);
		ans = ans.multiply(xx);
		ans = ans.mod(mod);

//		System.out.println(ans);

		BigInteger ans1 = new BigInteger("0");
		ans1 = aa.modPow(nn, mod);
		ans1 = ans1.subtract(new BigInteger("1")).mod(mod);
		ans1 = ans1.multiply((aaa.modInverse(mod)).mod(mod));
		ans1 = ans1.mod(mod);
		ans1 = ans1.multiply(bb);

//		System.out.println(ans1);

		ans = ans.add(ans1);
		ans = ans.mod(mod);

		System.out.println(ans);
	}

}
