Problem:
Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

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

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.

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

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define F first
#define S second
#define int long long
#define ull  unsigned long long
#define pb push_back

using namespace std;
using namespace __gnu_pbds;
typedef  tree< int , null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const int N = 500000;
vector < int > g[N];

main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //freopen("input.txt", "r", stdin);
    //freopen("ou tput.txt", "w", stdout);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= m; i++){
        int u, v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    sort(g[1].begin(), g[1].end());
    sort(g[n].begin(), g[n].end());
    int l1 = 0;
    int l2 = 0;
    while(l1 < g[1].size() && l2 < g[n].size()){
        if (g[1][l1] < g[n][l2]) l1++; else
        if (g[1][l1] > g[n][l2]) l2++; else{
            cout << "POSSIBLE";
            return 0;
        }
    }
    cout << "IMPOSSIBLE";
}