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.io.*;
import java.util.*;
import java.math.*;

public class D {
    public static void main(String[]args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(f.readLine());
        BigInteger a = new BigInteger(st.nextToken());
        BigInteger b = new BigInteger(st.nextToken());
        BigInteger n = new BigInteger(st.nextToken());
        BigInteger x = new BigInteger(st.nextToken());
        BigInteger mod = new BigInteger("1000000007");
        if (a.compareTo(BigInteger.ONE) == 0) {
            BigInteger ans = x.add(n.multiply(b));
            ans = ans.mod(mod);
            System.out.println(ans);
        } else {
            BigInteger ans = a.modPow(n, mod);
            ans = ans.multiply(x);
            BigInteger other = ((a.modPow(n, mod).subtract(BigInteger.ONE)).multiply((a.subtract(BigInteger.ONE)).modInverse(mod))).multiply(b);
            ans = ans.add(other);
            ans = ans.mod(mod);
            System.out.println(ans);
        }
    }
}