Longtail Hedgehog

Description

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 u_i and v_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.

Sample Input

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

Hint

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.

这道题需要求尾巴长度*刺的个数最大值，尾巴需要递增，刺是没有要求的，

就算尾巴和刺是同一个也没用关系。

只是处理不能用图来处理，直接对小的边排序，那么就可以求出，以某点

结尾的最长递增长度*与此点连接边的个数。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define LL long long
const int maxn = 100005;
int l[maxn], cnt[maxn];
struct node
{
    int u, v;
    node ( ) { }
    node ( int a, int b ) { u = a, v = b; }
    friend bool operator < ( node n1, node n2 )
    {
        return n1.u < n2.u;
    }
}a[maxn*4];
inline int Max ( int a, int b ) { return a > b ? a : b; }
int main ( )
{
    int n, m, u, v, pos = 0;
    LL ans = 0;
    scanf ( "%d%d", &n, &m );
    while ( m -- )
    {
        scanf ( "%d%d", &u, &v );
        if ( u > v ) //因为要求的是递增的序列,所以让小的在前面
        {
            int t = u;
            u = v;
            v = t;
        }
        cnt[v] ++;  //刺的个数(即与其相连的边的个数)
        cnt[u] ++;
        a[pos ++] = node ( u, v );
    }
    std :: sort ( a, a+pos );   //对第一个数排序,那样就可以求出尾巴的长度
    for ( int i = 0; i < pos; i ++ )
    {
        u = a[i].u, v = a[i].v;
        l[u] = Max ( l[u], 1 ); //小的有可能为0
        l[v] = Max ( l[v], l[u]+1 );    //直接等于小的+1或本身的个数
    }
    for ( int i = 1; i <= n; i ++ )
    {
        LL val = ( LL )l[i]*cnt[i];
        if ( ans < val )
            ans = val;
    }
    printf ( "%lld\n", ans );
    return 0;
}