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: Time Limit Exceeded)

#include<iostream>
#include <algorithm>
#include<string>
#include <bitset>
#include <vector>
#include <functional>
#include <climits>
#include <iomanip>
#include <utility>
#include <stack>
#include <queue>
#include <math.h>
#include <iomanip>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <numeric>
#include <stack>
#include <map> // pair
typedef std::stack<int> IntStack;
typedef std::stack<char> charStack;

using namespace std;
//using ll = long long;
using ll = unsigned long long;
typedef pair <ll, ll > pll;

int main()
{
	ll a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0, h = 0, ans = 0;
	ll n, m, K;
	ll memo[1003][1003] = { {} };
	ll memox[1003] = {};
	ll memoy[1003] = {};
	ll x[1000006] = {};
	ll y[200005] = {};
	ll z[100005] = {};
	string s;

	//IntStack yStack;
	//charStack Stack;

	cin >> a;
	cin >> b;
	for (int i = 0; i < b; i++)
	{
		cin >> x[i];
		cin >> y[i];
		if (x[i]>y[i])
		{
			c = x[i];
			x[i] = y[i];
			y[i] = c;
		}
	}
	for (int i = 0; i < b; i++)
	{
		if (x[i] == 1)
		{
			for (int k = 0; k < b; k++)
			{
				if (x[k] == y[i] && y[k] == a)
				{
					cout << "POSSIBLE" << endl;
					return 0;
				}
			}
		}
	}
	cout << "IMPOSSIBLE" << endl;


	//cout << a << endl;


	////std::cout << std::fixed;
	//cout << std::setprecision(20) << << endl;
	//cout << d+1 << endl;
	//cout << "" << endl;

	return 0;
}
//char型の文字数字は、-'0'で数値に変換できる
//accumulate(x, x + a, 0);//int型までの配列の合計
//memcpy(x, visited, sizeof(ll)*n);
//void kansuu(ll num, ll visited[10]) {}
//kansuu(1, zz);
//cout << setprecision(15) << v << endl;
//cout << char(z - 'a');
//log10(i) + 1 //10進数の桁数を取得

//int temp;
//for (int i = 0; i < n - 1; i++) {
//	for (int j = n - 1; j > i; j--) {
//		if (w[j - 1] < w[j]) {  /* 前の要素の方が大きかったら */
//			temp = w[j];        /* 交換する */
//			w[j] = w[j - 1];
//			w[j - 1] = temp;
//		}
//	}
//}

//最小公倍数
//ll x = a * b;
//ll tmp;
///* 自然数 a > b を確認・入替 */
//if (a<b) {
//	tmp = a;
//	a = b;
//	b = tmp;
//}
///* ユークリッドの互除法 */
//ll r = a % b;
//while (r != 0) {
//	a = b;
//	b = r;
//	r = a % b;
//}
//a = x / b;//答え