Problem:
Score : 800 points

Problem Statement

Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N).
Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence.
Then, perform the following operation until p becomes empty:

Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.

When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

N is an even number.

2 ≤ N ≤ 2 × 10^5

p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print the lexicographically smallest permutation, with spaces in between.

Sample Input 1

4
3 2 4 1

Sample Output 1

3 1 2 4

The solution above is obtained as follows:

p

q

(3, 2, 4, 1)

()

↓

↓

(3, 1)

(2, 4)

↓

↓

()

(3, 1, 2, 4)

Sample Input 2

2
1 2

Sample Output 2

1 2

Sample Input 3

8
4 6 3 2 8 5 7 1

Sample Output 3

3 1 2 7 4 6 8 5

The solution above is obtained as follows:

p

q

(4, 6, 3, 2, 8, 5, 7, 1)

()

↓

↓

(4, 6, 3, 2, 7, 1)

(8, 5)

↓

↓

(3, 2, 7, 1)

(4, 6, 8, 5)

↓

↓

(3, 1)

(2, 7, 4, 6, 8, 5)

↓

↓

()

(3, 1, 2, 7, 4, 6, 8, 5)

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)
#define pb push_back
#define f first
#define s second
/*
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll max3(ll a, ll b, ll c){return max(a, max(b, c));};
ll min3(ll a, ll b, ll c){return min(a, min(b, c));};
ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};
ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};
ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};
ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};
*/
const ll INFL = 1LL << 60;//10^18 = 2^60
const int INF = 1 << 30;//10^9
ll MOD = 1000000007;
//ll MOD = 998244353;

vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};

//デバッグ
void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
ll pow_mod(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){return x / gcd(x, y) * y;};

ll kai_mod(ll x){
  if(x == 0) return 1;
  return x * kai_mod(x-1) % MOD;
}

/*
//コンビネーション
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
ll comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
}
*/


ll N, M;
vector<ll> E[1010];
bool visited[1010];
bool ok;

void dfs(ll now, ll from, ll start){
  if(now == start && from != -1){
    ok = 1;
    return;
  }
  if(visited[now]) return;
  visited[now] = 1;
  for(ll to : E[now]){
    dfs(to, now, start);
  }
}

ll dist[1010];
ll last;

void dijkstra(ll s){
  rep(i, 1010) dist[i] = INFL;
  dist[s] = 0;
  priority_queue<P, vector<P>, greater<P>> pque;
  pque.push({0, s});
  while(!pque.empty()){
    P p = pque.top();
    pque.pop();
    if(dist[p.s] < p.f) continue;
    for(ll to : E[p.s]){
      if(to == s){
        last = p.s;
        continue;
      }
      if(dist[to] > p.f + 1){
        dist[to] = p.f + 1;
        pque.push({dist[to], to});
      }
    }
  }
}


pair<ll, pair<ll, ll>> solve(ll start){
  dijkstra(start);
  return {dist[last], {last, start}};

}


int main(){
  cin >> N >> M;
  rep(i, M){
    ll a, b;
    cin >> a >> b;
    a--, b--;
    E[a].pb(b);
  }
  vector<pair<ll, pair<ll, ll>>> ANS;
  rep(i, N){
    ok = 0;
    rep(j, N) visited[j] = 0;
    dfs(i, -1, i);
    if(ok){
      ANS.pb(solve(i));
    }
  }
  if(ANS.size() == 0){
    cout << -1 << endl;
    return 0;
  }
  sort(ANS.begin(), ANS.end());

  rep(a, ANS.size()){
    dijkstra(ANS[a].s.s);
    set<ll> ans;
    ll now = ANS[a].s.f;
    while(true){
      ans.insert(now);
      if(now == ANS[a].s.s) break;
      ll d = dist[now];
      rep(i, 1010){
        if(dist[i] == d - 1){
          now = i;
        }
      }
    }
    ll to_cnt[1010], in_cnt[1010];
    rep(i, 1010){
      to_cnt[i] = 0;
      in_cnt[i] = 0;
    }
    for(auto itr = ans.begin(); itr != ans.end(); itr++){
      ll now = *itr;
      for(ll to : E[now]){
        if(ans.count(to) == 0) continue;
        to_cnt[now]++;
        in_cnt[to]++;
      }
    }
    bool p = 1;
    rep(i, 1010){
      if(in_cnt[i] > 1) p = 0;
      if(to_cnt[i] > 1) p = 0;
    }

    if(p == 0) continue;
    cout << ans.size() << endl;
    for(auto itr = ans.begin(); itr != ans.end(); itr++){
      cout << *itr + 1 << endl;
    }
    return 0;
  }
  cout << -1 << endl;
}