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 CPP):

Solution:
#include <bits/stdc++.h>
const int oo = INT_MAX;
using namespace std;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int dx1[] = {-1, 1, 0, 0};
int dy1[] = {0, 0, -1, 1};
struct dsu {
  vector<int> level, par;
  int forest;
  dsu(int n) {
    par = level = vector<int>(n);
    forest = n;
    for (int i = 0; i < n; ++i) level[i] = 1, par[i] = i;
  }
  int find_dsu(int node) {
    if (par[node] == node) return node;
    return par[node] = find_dsu(par[node]);
  }
  void link_dsu(int node_1, int node_2) {
    if (level[node_2] > level[node_1])
      swap(node_1, node_2);
    else if (level[node_1] == level[node_2])
      level[node_1]++;
    par[node_2] = node_1;
  }
  bool union_dsu(int node_1, int node_2) {
    node_1 = find_dsu(node_1), node_2 = find_dsu(node_2);
    if (node_1 != node_2) {
      link_dsu(node_1, node_2);
      forest--;
      return true;
    }
    return false;
  }
  bool same_set(int node_1, int node_2) {
    node_1 = find_dsu(node_1), node_2 = find_dsu(node_2);
    return node_1 == node_2;
  }
};
struct edge {
  int from, to;
  double w;
  edge(int from, int to, double w) : from(from), to(to), w(w){};
  bool operator<(const edge& e) const { return w < e.w; }
};
double legnth(pair<double, double> a, pair<double, double> b) {
  return (((a.first - b.first) * (a.first - b.first)) +
          ((a.second - b.second) * (a.second - b.second)));
}
float mst(vector<edge> edgelist, int n) {
  dsu un(n);
  for (auto i : edgelist) {
    un.union_dsu(i.from, i.to);
    if (un.same_set(0, 1)) return i.w;
  }
}
void file() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  ;
}
int main() {
  file();
  long long n, m, v;
  cin >> n >> m >> v;
  int x = -1;
  if (m >= n - 1) {
    for (int i = 1; n - 1 - i > 0; ++i) {
      if ((long long)(i + 1) * i / 2 + (long long)(n - i) * (n - 1 - i) / 2 >=
          m) {
        x = i;
        break;
      }
    }
  }
  if (x == -1)
    cout << -1 << '\n';
  else {
    int y = n - 1 - x;
    int id[100005];
    for (int i = 1, j = 0; i <= n; ++i)
      if (i != v) id[j++] = i;
    for (int i = 0; i + 1 < x; ++i) cout << id[i] << " " << id[i + 1] << '\n';
    cout << id[x - 1] << " " << v << '\n';
    for (int i = 0; i + 1 < y; ++i)
      cout << id[x + i] << " " << id[x + i + 1] << '\n';
    cout << id[n - 2] << " " << v << '\n';
    m -= n - 1;
    for (int i = 0; i < x && m > 0; ++i)
      for (int j = i + 2; j < x && m > 0; ++j) {
        cout << id[i] << " " << id[j] << '\n';
        --m;
      }
    for (int i = 0; i < y && m > 0; ++i)
      for (int j = i + 2; j < y && m > 0; ++j) {
        cout << id[x + i] << " " << id[x + j] << '\n';
        --m;
      }
    for (int i = 0; i < x - 1 && m > 0; ++i) {
      cout << id[i] << " " << v << '\n';
      --m;
    }
    for (int i = 0; i < y - 1 && m > 0; ++i) {
      cout << id[x + i] << " " << v << '\n';
      --m;
    }
  }
}