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 <iostream>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<set>
#define LL long long
#define INF 100000009
#define maxn 1000009
#define mod 1000000007
#include <set>
#include<iostream>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
int vis[200001];
int main()
{
    int n,m,a,b;
    scanf("%d %d",&n,&m);
    memset(vis,0,sizeof(vis));
    for(int i=0;i<m;i++)
    {
        scanf("%d %d",&a,&b);
        if(a==1||b==1)
        {
            if(a==1)
            vis[b]++;
            else
            vis[a]++;
        }
        if(a==n||b==n)
        {
            if(a==n)
            vis[b]++;
            else
            vis[a]++;
        }
    }
    int flag=0;
    for(int i=0;i<=n;i++)
    {
        if(vis[i]==2)
        flag=1;
    }
    if(flag)
    printf("POSSIBLE\n");
    else
    printf("IMPOSSIBLE\n");
}