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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define si(X) scanf("%d", &(X))
#define sll(X) scanf("%lld",&(X))
#define INFL 0x3f3f3f3f3f3f3f3fLL
typedef vector<vector<ll> > matrix;
const int mod = 1e9+7;
ll gcd(ll a,ll b){
	if(b==0)return a;return gcd(b,a%b);
}
ll expo(ll base,ll pow){
    ll ans = 1;
    while(pow!=0){
        if(pow&1==1){ans = ans*base;ans = ans%mod;}
        base *= base;base%=mod;pow/=2;}return ans;
}
ll inv(ll x){return expo(x,mod-2);}

double pi = 3.141592653589793238462643;
double error = 0.0000001;
int dx[8] = {1 , 0 , -1 , 0 , 1 , -1 , -1 , 1};    // last 4 diagonal
int dy[8] = {0 , 1 , 0 , -1 , 1 , 1 , -1 , -1};
/* -------Template ends here-------- */

const int M = 210001;

vector<int> vec[M];
int vis[M];

void dfs(int u){
    vis[u] = 1;
    for(int i = 0 ; i < vec[u].size() ; i++){
        int v = vec[u][i];
        if(vis[v]) continue;
        dfs(v);
    }
}

int main(){
    //freopen("input.txt", "rt", stdin);
    //freopen("output.txt", "wt", stdout);


    int n , m;
    si(n);  si(m);

    for(int i = 1 ; i <= m ; i++){
        int u , v;
        si(u);  si(v);
        vec[u].push_back(v);
        vec[v].push_back(u);
    }

    dfs(1);
    if(vis[n]){
        cout<<"POSSIBLE";
    }
    else{
        cout<<"IMPOSSIBLE";
    }


















}