长尾狐狸（简单dp）

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 of n 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 the endpoint 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 and m(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 integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — 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 64 53 52 51 22 86 7

output

9

input

4 61 21 31 42 32 43 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 numbers 1, 2 and 5. The following segments are spines: (2, 5), (3, 5) and (4, 5). Therefore, the beauty of the hedgehog is equal to 3·3 = 9.

题意：

找图中最长递增序列与末尾点度乘积的最大值。

分析：

最长递增序列的求法：要求最长递增序列的最后一个点的dp值，则必须知道它的前一个点的递增序列长度。

dp[i]，i表示点，dp[i]表示末尾点为i时的递增序列长度。

图的存储：一个一维数组存储每个点的度。

vector[MAXN]存储每个点的边。

启发：对于每道题，要善于发现数与数之间的规律，不只是数会表现出一定的特点，其他的也可能会，所以一定要善于发现。

//main.cpp 
//XUPT08-2 长尾狐狸 
//
//Created BY YOUMI on 16/9/26
//
//简单dp，图用vector数组存储
// 
 
#include<iostream> 
#include<stdio.h>
#include<algorithm>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<cstdlib>
#include<utility>
using namespace std;
const int MAXN=1e5+5;
int n,m;
int dp[MAXN];
int du[MAXN];
vector<int>edge[MAXN];
long long int ans;




int main(int argc,const char* argv[])
{
scanf("%d%d",&n,&m);
int a,b;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
if(a>b) swap(a,b);
du[a]++;
du[b]++;
edge[b].push_back(a);
}

ans=0;
for(int i=1;i<=n;i++)
{
dp[i]=1;
for(int j=0;j<edge[i].size();j++)
{
dp[i]=max(dp[i],dp[edge[i][j]]+1);
}
ans=max(ans,1ll*du[i]*dp[i]);
}
printf("%d\n",ans);
return 0;
}