// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long m,n,i,j,x,y,z,a[300000],k=0,cnt=0,b[300000];

    cin>>n>>m;

    memset(a,0,sizeof(a));

    for(i=0;i<m;i++)

    {

        cin>>x>>y;

        a[x]=y;

        if(x==1)

        {

            b[k]=y;

            k++;

        }

    }

    for(i=0;i<k;i++)

    {

        if(a[b[i]]==n)

        {

            cnt=1;

            break;

        }

    }

    if(cnt==1)

        cout<<"POSSIBLE"<<endl;

    else

        cout<<"IMPOSSIBLE"<<endl;





}


// Faster C++ solution for the same problem (measured speedup: 3.501342373383471x):
#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/pb_ds/detail/standard_policies.hpp>



using namespace std;

using namespace __gnu_pbds;

long long power(long long x, long long y, long long mood) {

	if (y == 0)

		return 1;

	long long ret = power(x, y >> 1, mood) % mood;

	if (y % 2 == 0)

		return (ret * ret) % mood;

	else

		return (x * (ret * ret % mood)) % mood;

}

const double Eps = (1e-9);

int dcmp(double x, double y) {

	return fabs(x - y) <= Eps ? 0 : x < y ? -1 : 1;

}

typedef tree<int, null_type, less<int>, rb_tree_tag,

		tree_order_statistics_node_update> ordered_set;

typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag,

		tree_order_statistics_node_update> ordered_setpair;

typedef long long ll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<string> vs;

typedef vector<pair<int, int> > vii;

typedef map<int, int> mii;

typedef map<string, int> msi;

typedef pair<int, int> pii;

#define mii map <int, int>

#define mll map <ll, ll>

#define pii pair <int, int>

#define pll pair <ll, ll>

#define vi vector <int>

#define vd vector <double>

#define vll vector <ll>

#define sll set <ll>

#define spii set <pii>

#define vpii vector <pair <int,int> >

#define vpll vector <pair <long long,long long> >

#define vvi vector <vector <int> >

#define vvpii vector <vector <pii > >

#define sfi1(v) scanf("%d",&v)

#define sfi2(v1,v2) scanf("%d %d",&v1,&v2)

#define sfi3(v1,v2,v3) scanf("%d %d %d",&v1,&v2,&v3)

#define sfll1(v) scanf("%I64d",&v);

#define sfll2(v1,v2) scanf("%I64d %I64d",&v1,&v2)

#define sfll3(v1,v2,v3) scanf("%I64d %I64d %I64d",&v1,&v2,&v3)

#define min3(a,b,c)     min(a,min(b,c))

#define max3(a,b,c)     max(a,max(b,c))

#define min4(a,b,c,d)   min(a,min(b,min(c,d)))

#define max4(a,b,c,d)   max(a,max(b,max(c,d)))

#define el	'\n'

#define lp(i,n) for(int i = 0;i < int(n);i++)

#define lp1(i,n) for(int i = 1;i <= int(n);i++)

#define rlp(i, n) for (int i = (n) - 1; i >= 0; i--)

#define rng(i,a,b) for(int i = int(a);i <= int(b);i++)

#define mp(x,y)  	make_pair((x),(y))

#define pb(x)    	push_back((x))

#define pob()    	pop_back()

#define all(v)  	((v).begin()),((v).end())

#define sz(x)  		((int) (x).size())

#define bits(a) 	__builtin_popcountll(a)

#define fr first

#define sc second

#define pii pair<int, int>

#define sortva(v) sort(all(v))

#define sortvd(v) sort(v.rbegin(),v.rend())

#define sortaa(a,n) sort(a,a+n)

#define sortad(a,n) sort(a,a+n),reverse(a,a+n)

#define SS stringstream

#define ll long long

#define clr(x, y) memset(x, y, sizeof(x))

#define foreach(it,x) for(__typeof__((x).begin()) it = (x).begin(); it != (x).end(); ++it)

#define read freopen("input.in", "rt", stdin);

#define write freopen("output.in", "wt", stdout);

#define fastIO cout << fixed << setprecision(7), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

const long long key1 = (31);

const long long key2 = 59;

int dx[] = { 0, 1, 0, -1, 1, -1, 1, -1 };

int dy[] = { 1, 0, -1, 0, -1, 1, 1, -1 };

int kx[] = { 2, 2, 1, -1, -2, -2, 1, -1 };

int ky[] = { 1, -1, 2, 2, 1, -1, -2, -2 };

int gcd(int a, int b) {

	if (a == 0)

		return b;

	return gcd(b % a, a);

}

ll lcm(ll a, ll b) {

	return (a / gcd(a, b)) * b;

}

int setBit(int num, int idx, int val) {

	return val ? (num | (1 << idx)) : (num & (~(1 << idx)));

}

int getBit(int num, int idx) {

	return ((num >> idx) & 1) == 1;

}

int const N = 1e6+5 , mod = 1e9 + 7, oo = 1e9 + 10;

//-------------------------------------------------------------------------------------

int arr[N];

int n;

int one[N];

int last[N];

int main()

{

	int n,m;

	cin>>n>>m;

	lp(i,m){

	int x,y;

	cin>>x>>y;

	if (x==n)last[y]=1;

	if (y==n)last[x]=1;

	if (x==1)one[y]=1;

	if (y==1)one[x]=1;

	}

	for (int i=2;i<n;i++)if (one[i]&& last[ i ])return cout<< "POSSIBLE",0;

	cout<< "IMPOSSIBLE";













	return 0;

}