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

/**
 *
 * @author sarthak
 */

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

public class edu13_D {

    static class FastScanner {

        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public String nextLine() {
            st = null;
            try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
                return "";
            }
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
    static long mod = 1000000000 + 7;

    public static long pw(long a, long n) {
        if (n == 0) {
            return 1;
        }
        long h = pw(a, n / 2);
        h = ((h % mod) * (h % mod)) % mod;
        if (n % 2 != 0) {
            h = ((h % mod) * (a % mod)) % mod;
        }
        return h;
    }

    public static void main(String[] args) {

        FastScanner s = new FastScanner(System.in);
        StringBuilder op = new StringBuilder();

        long A = s.nextLong();
        long B = s.nextLong();
        long n = s.nextLong();
        long x = s.nextLong();
        long v = pw(A, n);
        if(A==1){
            long an=(x%mod + ((B%mod)*(n%mod))%mod)%mod;
            System.out.println(an);
            return;
        }
        else{
            BigInteger a=BigInteger.valueOf(A-1);
            long w=a.modInverse(BigInteger.valueOf(mod)).longValue();
            w=((w%mod)*(B%mod))%mod;
            w=((w%mod)*((v-1)%mod))%mod;
            long an=(((v%mod)*(x%mod))%mod +w%mod)%mod;
            System.out.println(an);
                        
        }
        
    }

}
