The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She'll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1.

Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she's only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.

Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she'll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she'll have to traverse.

Input

* Line 1: Two space-separated integers, N and M.

* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.

Output

* Line 1: A single integer that is the length of the longest road required to be traversed.

Sample Input

3 3
1 2 23
2 3 1000
1 3 43

Sample Output

43

Hint

OUTPUT DETAILS:

In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.

題意：

小女孩想從1到達N， 求小女孩到達目的地邊的最小路徑的最大邊。

一開始最先想到的是迪傑斯特拉演算法， 飛速寫了個然後光速去世。 然後搜了下， 發現原來是最小生成樹。 。。感覺自己好蠢。

。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=10005;
const int max1=2005;
struct edge
{
    int s,e;
    int sp;
};
edge e[maxn];
int n,m;
int a[max1];
int compare (edge a,edge b)
{
    return a.sp<b.sp;
}
void init ()
{
    for (int i=1;i<=n;i++)
        a[i]=i;
}
int finds (int x)
{
    if(a[x]==x)
        return x;
    return a[x]=finds(a[x]);
}
bool unit(int x,int y)
{
    int temp1=finds(x);
    int temp2=finds(y);
    if(temp1!=temp2)
    {
        a[temp1]=temp2;
        return true;
    }
    return false;
}
void kual ()
{
    sort (e,e+m,compare);
    int num=n,maxx=-1;
    for (int i=0;i<m&&num>1;i++)
    {
        if(unit(e[i].s,e[i].e))
        {
            num--;
            maxx=max(maxx,e[i].sp);
        }
    }
    printf("%d\n",maxx);
}
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        for (int i=0;i<m;i++)
            scanf("%d%d%d",&e[i].s,&e[i].e,&e[i].sp);
        kual();
    }
    return 0;
}