Codeforces--615B--Longtail Hedgehog（贪心模拟）



B. Longtail Hedgehog

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

This Christmas Santa gave Masha a magic picture and a pencil. The picture consists ofn points connected by
m segments (they might cross in any way, that doesn't matter). No two segments connect the same pair of points, and no segment connects the point to itself. Masha wants to color some segments in order paint a hedgehog.
In Mashas mind every hedgehog consists of a tail and some spines. She wants to paint the tail that satisfies the following conditions:

1. Only segments already presented on the picture can be painted;
2. The tail should be continuous, i.e. consists of some sequence of points, such that every two neighbouring points are connected by a colored segment;
3. The numbers of points from the beginning of the tail to the end should strictly increase.

Masha defines the length of the tail as the number of points in it. Also, she wants to paint some spines. To do so, Masha will paint all the segments, such that one of their ends is theendpoint of the tail. Masha defines
the beauty of a hedgehog as the length of the tail multiplied by the number of spines. Masha wants to color the most beautiful hedgehog. Help her calculate what result she may hope to get.

Note that according to Masha's definition of a hedgehog, one segment may simultaneously serve as a spine and a part of the tail (she is a little girl after all). Take a look at the picture for further clarifications.

Input

First line of the input contains two integers n andm(2 ≤ n ≤ 100 000,1 ≤ m ≤ 200 000) — the number
of points and the number segments on the picture respectively.

Then follow m lines, each containing two integersu_i andv_i
(1 ≤ u_i, v_i ≤ n,u_i ≠ v_i) —
the numbers of points connected by corresponding segment. It's guaranteed that no two segments connect the same pair of points.

Output

Print the maximum possible value of the hedgehog's beauty.

Examples

Input

8 6
4 5
3 5
2 5
1 2
2 8
6 7

Output

9

Input

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

Output

12

Note

The picture below corresponds to the first sample. Segments that form the hedgehog are painted red. The tail consists of a sequence of points with numbers1,
2 and 5. The following segments are spines: (2,
5), (3,
5) and (4,5). Therefore, the beauty of the hedgehog is equal to3·3 = 9.

n个点，m条无向边，在连成的链中找一条递增的链，使得末尾节点的度数乘以深度最大，因为是无向边，又要求递增的链，所以尽量使小的数做起点，并且将所有的边按照节点大小进行排序，从最小的点开始遍历，记录每一个点最大的深度，然后找到最大的乘积

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
	int u,v;
}edge[200000+10];
bool cmp(node s1,node s2)
{
	if(s1.u==s2.u)
	return s1.v<s2.v;
	return s1.u<s2.u;
}
int main()
{
	int n,m;
	cin>>n>>m;
	__int64 dp[100100],dre[100100];
	memset(dp,0,sizeof(dp));
	memset(dre,0,sizeof(dre));
	for(int i=0;i<m;i++)
	{
		cin>>edge[i].u>>edge[i].v;
		if(edge[i].u>edge[i].v)
		swap(edge[i].u,edge[i].v);//¾¡Á¿Ê¹Ð¡µÄµã×öÆðµã
		dre[edge[i].u]++,dre[edge[i].v]++;
	}
	sort(edge,edge+m,cmp);
	for(int i=0;i<m;i++)//¼Ç¼ÿһ¸öµãµÄ×î´óÉî¶È
	dp[edge[i].v]=max(dp[edge[i].v],dp[edge[i].u]+1);
	__int64 ans=0;
	for(int i=1;i<=n;i++)
	ans=max(ans,(dp[i]+1)*dre[i]);//Ñ°ÕÒ×î´ó³Ë»ý
	cout<<ans<<endl;
	return 0;
}