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>
using namespace std;
inline int inp() { return 0; }
inline int inp(bool& A) {
  if (cin >> A) return 1;
  return -1;
}
inline int inp(char& A) { return scanf(" %c", &A); }
inline int inp(int& A) { return scanf("%d", &A); }
inline int inp(float& A) {
  if (cin >> A) return 1;
  return -1;
}
inline int inp(long long& A) { return scanf("%I64d", &A); }
inline int inp(double& A) { return scanf("%lf", &A); }
inline int inp(char* A) { return scanf("%s", A); }
inline int inp(string& A) {
  if (cin >> A) return 1;
  return -1;
}
template <class Front, class... Queue>
inline int inp(Front& A, Queue&... B) {
  return inp(A) + inp(B...);
}
inline int oup() { return 0; }
inline int oup(bool A) {
  if (cout << A) return 1;
  return -1;
}
inline int oup(char A) { return printf("%c", A); }
inline int oup(int A) { return printf("%d", A); }
inline int oup(float A) {
  if (cout << A) return 1;
  return -1;
}
inline int oup(long long A) { return printf("%I64d", A); }
inline int oup(double A) { return printf("%lf", A); }
inline int oup(char* A) { return printf(A); }
inline int oup(const char* A) { return printf("%s", A); }
inline int oup(string& A) {
  if (cout << A) return 1;
  return -1;
}
template <class Front, class... Queue>
inline int oup(Front A, Queue... B) {
  return oup(A) + oup(B...);
}
template <class T>
inline void remax(T& A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  for (int i = s.size() - 1; i >= 0; --i) r += (s[i] - '0') * p, p *= 10;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, long long power) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
long long PowerMod(long long base, long long power, long long mod) {
  if (!power) return 1;
  if (power & 1) return (base * PowerMod(base, power - 1, mod)) % mod;
  return PowerMod((base * base) % mod, power >> 1, mod);
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
int Count(long long mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
inline void run() {
in:;
  int n, m, v;
  if (inp(n, m, v) != 3) return;
  if (m < n - 1) {
    oup("-1\n");
    return;
  }
  if (m > (((n - 1) * (n - 2)) >> 1) + 1) {
    oup("-1\n");
    return;
  }
  int tmp = 1;
  if (v == 1) ++tmp;
  for (int i = 1; i <= n; ++i)
    if (i != v) oup(i, ' ', v, '\n');
  m -= n - 1;
  for (int i = 1; i <= n && m; ++i)
    if (i != v && i != tmp)
      for (int j = i + 1; j <= n && m; ++j)
        if (j != v && j != tmp) {
          oup(i, ' ', j, '\n');
          --m;
        }
  goto in;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  run();
  return 0;
}