Codeforces Round #562 (Div. 2) B - Pairs

B. Pairs

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Toad Ivan has m

pairs of integers, each integer is between 1 and n, inclusive. The pairs are (a1,b1),(a2,b2),…,(am,bm)

.

He asks you to check if there exist two integers x

and y (1≤x<y≤n) such that in each given pair at least one integer is equal to x or y

.

Input

The first line contains two space-separated integers n

and m (2≤n≤300000, 1≤m≤300000

) — the upper bound on the values of integers in the pairs, and the number of given pairs.

The next m

lines contain two integers each, the i-th of them contains two space-separated integers ai and bi (1≤ai,bi≤n,ai≠bi) — the integers in the i

-th pair.

Output

Output "YES" if there exist two integers x

and y (1≤x<y≤n) such that in each given pair at least one integer is equal to x or y

. Otherwise, print "NO". You can print each letter in any case (upper or lower).

Examples

Input

Copy

4 6
1 2
1 3
1 4
2 3
2 4
3 4

Output

Copy

NO

Input

Copy

5 4
1 2
2 3
3 4
4 5

Output

Copy

YES

Input

Copy

300000 5
1 2
1 2
1 2
1 2
1 2

Output

Copy

YES

Note

In the first example, you can't choose any x

, y

because for each such pair you can find a given pair where both numbers are different from chosen integers.

In the second example, you can choose x=2

and y=4

.

In the third example, you can choose x=1

and y=2

.

思路：我的思路是把每个数的所有位置分别存到1个set里去，然后判断有没有两个数去掉重复（用set.find）位置后位置数和等于n。。。这个思路写了我快一个小时，实际上以第一个数对中的x，y为基准，把其他数对中含有x的去掉，剩下的数对如果都含有y，那就输出YES，否则输出NO就好了

 By jswwsj, contest: Codeforces Round #562 (Div. 2), problem: (B) Pairs, Accepted, #

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<set>
#define ll long long
#define mod 1000000007
using namespace std;
typedef struct Dt
{
	set<int>q;
}Dt;
Dt a[300050],b[305000];
bool cmp(Dt x,Dt y)
{
    return x.q.size()>y.q.size();
}
int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=m;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            a[x].q.insert(i);
            a[y].q.insert(i);
        }
        int t=0;
        for(int i=1;i<=n;i++)
        {
            if(!a[i].q.empty())
            {
                a[i].q.swap(b[t].q);
                //printf("i=%d b.[%d].size=%d\n",i,t,b[t].q.size());
                t++;

            }
        }
        sort(b,b+t,cmp);
        int flag=0;
        for(int i=0;i<t;i++)
        {
            if(b[i].q.size()+b[i+1].q.size()<m) break;
            for(int j=i+1;j<t;j++)
            {
                Dt x;
                int sum=b[i].q.size()+b[j].q.size();
                while(!b[j].q.empty())
                {
                    int m;
                    m=*b[j].q.begin();
                    b[j].q.erase(m);
                    x.q.insert(m);
                    if(b[i].q.find(m)!=b[i].q.end()) sum--;
                }
                b[j].q.swap(x.q);
                //printf("b[%d].size=%d b[%d].size=%d sum=%d\n",i,b[i].q.size(),j,b[j].q.size(),sum);
                if(sum>=m)
                {
                    flag=1;
                    break;
                }

            }
            if(flag==1) break;
        }
        if(flag==1) printf("YES\n");
        else printf("NO\n");
    }
}