題干：

Do you know a story about the three musketeers? Anyway, you will learn about its origins now.

Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.

There are?n?warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his?recognition?is the number of warriors he knows, excluding other two musketeers.

Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions.

Input

The first line contains two space-separated integers,?n?and?m?(3?≤?n?≤?4000,?0?≤?m?≤?4000) — respectively number of warriors and number of pairs of warriors knowing each other.

i-th of the following?m?lines contains two space-separated integers?ai?and?bi?(1?≤?ai,?bi?≤?n,?ai?≠?bi). Warriors?ai?and?bi?know each other. Each pair of warriors will be listed at most once.

Output

If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes).

Examples

Input

5 6 1 2 1 3 2 3 2 4 3 4 4 5

Output

2

Input

7 4 2 1 3 6 5 1 1 7

Output

-1

Note

In the first sample Richelimakieu should choose a triple?1,?2,?3. The first musketeer doesn't know anyone except other two musketeers so his recognition is?0. The second musketeer has recognition?1?because he knows warrior number?4. The third musketeer also has recognition?1?because he knows warrior?4. Sum of recognitions is?0?+?1?+?1?=?2.

The other possible triple is?2,?3,?4?but it has greater sum of recognitions, equal to?1?+?1?+?1?=?3.

In the second sample there is no triple of warriors knowing each other.

題目大意：

給你n個人和m個關系，(a,b)表示a和b認識，現在這個人想雇傭其中3個人，

1： 這3個人必須相互認識

2： 這3個人的識別度的總和最小（一個人的識別度為：除了另外兩人認識的人的數，三個人的識別度相加最小）

解題報告：

? ? 顯然，2000ms的時限，4000的數據量，考慮是否o(n^2)可以解決。（其實這題直接枚舉三個點，o(n^3)也可以過？）怎么優化到平方級別呢？其實也很簡單，就是第一層循環遍歷m條邊就好了。因為你三層for枚舉頂點的時候，其實前兩層有很多是浪費掉了的，因為有很多頂點的之間根本沒有邊，而你都遍歷了一遍，所以我們都記錄下來邊了，為什么不直接遍歷邊呢？這樣我們就有兩個頂點了呀，然后再來一層循環遍歷每一個點就好了。注意這種暴力的題一定別忘判斷一下是否遍歷到了自己，比如【CodeForces - 761C】Dasha and Password 這道題，最開始就忘了判斷是否遍歷到了同一個字符串。

AC代碼：

#include<bits/stdc++.h>using namespace std; const int MAX = 4000 + 5 ; int du[MAX]; bool maze[MAX][MAX]; int head[MAX]; struct Edge {int fm,to;int w; }e[50000 + 5]; int top; void add(int u,int v) {e[++top].fm = u;e[top].to = v; } int main() {int n,m,u,v;cin>>n>>m;for(int i = 1; i<=m; i++) {scanf("%d%d",&u,&v);add(u,v);add(v,u);du[u]++;du[v]++;maze[u][v] = maze[v][u] = 1;}int ans = 0x3f3f3f3f;for(int i = 1; i<=top; i++) {for(int j = 1; j<=n; j++) {if(j == e[i].fm || j == e[i].to) continue;if(maze[e[i].fm][j] && maze[e[i].to][j]) {ans = min(ans,du[e[i].to] + du[e[i].fm] + du[j]);}}}if(ans == 0x3f3f3f3f) puts("-1");else printf("%d\n",ans-6);return 0 ;}

再來一個代碼：

#include<bits/stdc++.h> using namespace std; #define ll long long const int inf = 0x3f3f3f3f; const int maxn = 4050;int n,m,d[maxn]; bool g[maxn][maxn]; //一開始以為暴力的話是n^3會超時，后來看了題解原來前兩層放上去之后加個剪枝，第三層的復雜度是加上去的而不是乘上去的 //即剪枝使得復雜度由O(N^3)降為O(N^2+NM) int main(){//62 ms 18100 KBint x,y;scanf("%d%d",&n,&m);for(int i=1;i<=m;++i){scanf("%d%d",&x,&y);g[x][y]=g[y][x]=1;d[x]++;d[y]++; }int ans=inf;for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){if(g[i][j] && d[i]+d[j]<=inf){//只有M個g[i][j]成立 for(int k=1;k<=n;++k){if(g[i][k] && g[j][k]){ans=min(ans,d[i]+d[j]+d[k]);}}}}}//相當于外面枚舉N^2次，對于其中的M次才有第三層for 故復雜度為 O(N^2+M*N) if(ans>=inf) puts("-1");else printf("%d\n",ans-6);return 0; }

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AC代碼2：（其實也不是拓撲排序啦，就是用vector表示了二維數組，并且，由第二個點找第三個點的時候直接在第一個點所連的邊中找就可以了。但是那個c函數其實也有點消耗時間？如果直接用個bool類型的maze二維數組來表示無向圖兩點之間是否相連就更快了）

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總結

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