Problem: Bob got a job as a system administrator in X corporation. His first task was to connect n servers with the help of m two-way direct connection so that it becomes possible to transmit data from one server to any other server via these connections. Each direct connection has to link two different servers, each pair of servers should have at most one direct connection. Y corporation, a business rival of X corporation, made Bob an offer that he couldn't refuse: Bob was asked to connect the servers in such a way, that when server with index v fails, the transmission of data between some other two servers becomes impossible, i.e. the system stops being connected. Help Bob connect the servers.

Input

The first input line contains 3 space-separated integer numbers n, m, v (3 ≤ n ≤ 105, 0 ≤ m ≤ 105, 1 ≤ v ≤ n), n — amount of servers, m — amount of direct connections, v — index of the server that fails and leads to the failure of the whole system.

Output

If it is impossible to connect the servers in the required way, output -1. Otherwise output m lines with 2 numbers each — description of all the direct connections in the system. Each direct connection is described by two numbers — indexes of two servers, linked by this direct connection. The servers are numbered from 1. If the answer is not unique, output any.

Examples

Input

5 6 3


Output

1 2
2 3
3 4
4 5
1 3
3 5


Input

6 100 1


Output

-1

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class C0022 {

    public static void main(String args[]) throws Exception {
        new C0022();
    }

    C0022() throws Exception {
        PandaScanner sc = null;
        PrintWriter out = null;
        try {
            sc = new PandaScanner(System.in);
            out = new PrintWriter(System.out);
        } catch (Exception ignored) {
        }

        int n = sc.nextInt();
        int m = sc.nextInt();
        int v = sc.nextInt();
        int singledOut = n;
        if (v == n) {
            singledOut--;
        }

        //Check for tree and overflow
        int tree = n - 1;
        long max = (((long) (n - 1) * (long) (n - 2)) >> 1) + 1;
        if (tree > m || m > max) {
            out.println("-1");
        }
        else {
            //Build a tree from v to everyone else
            for (int i = 1; i < n + 1 && m != 0; i++) {
                if (i != v) {
                    out.println(i + " " + v);
                    m--;
                }
            }
            //Use the rest of the edges
            for (int i = 1; i < n + 1 && m != 0; i++) {
                if (i == v || i == singledOut) {
                    continue;
                }
                for (int j = i + 1; j < n + 1 && m != 0; j++) {
                    if (j == v || j == singledOut) {
                        continue;
                    }
                    out.println(i + " " + j);
                    m--;
                }
            }

        }

        out.close();
        System.exit(0);
    }


    //The PandaScanner class, for Panda fast scanning!
    public class PandaScanner {
        BufferedReader br;
        StringTokenizer st;
        InputStream in;

        PandaScanner(InputStream in) throws Exception {
            br = new BufferedReader(new InputStreamReader(this.in = in));
        }

        public String next() throws Exception {
            if (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine().trim());
                return next();
            }
            return st.nextToken();
        }

        public boolean hasNext() throws Exception {
            return (st != null && st.hasMoreTokens()) || in.available() > 0;
        }

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

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