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 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 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：给你一个图 求 ans的最大值 ans=图中一条点权严格递增的链的长度乘以终点的度

题解：dp[i] 为以i为终点的满足条件的链的长度  并且计算每个点的度    枚举乘积取max输出

 1 #pragma comment(linker, "/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <cmath>
 8 #include <cctype>
 9 #include <map>
10 #include <set>
11 #include <queue>
12 #include <bitset>
13 #include <string>
14 #include <complex>
15 #define ll __int64
16 using namespace std;
17 int n,m;
18 map<int,ll>du;
19 struct node
20 {
21     int to;
22     int pre;
23 }N[400005];
24 int pre[400005];
25 int nedge=0;
26 void add(int s,int t)
27 {
28     nedge++;
29     N[nedge].to=t;
30     N[nedge].pre=pre[s];
31     pre[s]=nedge;
32 }
33 ll dp[100005];
34 int main()
35 {
36     scanf("%d %d",&n,&m);
37     int x,y;
38     du.clear();
39     for(int i=1;i<=n;i++){
40         dp[i]=1;
41         }
42     for(int i=1;i<=m;i++){
43         scanf("%d %d",&x,&y);
44         add(x,y);
45         add(y,x);
46         du[x]++;
47         du[y]++;
48     }
49     for(int i=2;i<=n;i++)
50     {
51        for(int j=pre[i];j;j=N[j].pre)
52        {
53            if(N[j].to<i)
54            {
55                dp[i]=max(dp[i],dp[N[j].to]+1);
56            }
57        }
58     }
59     ll ans=0;
60     for(int i=1;i<=n;i++){
61         ans=max(ans,du[i]*dp[i]);
62         }
63     printf("%I64d\n",ans);
64     return 0;
65 }