One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

題解：

第一個數字代表有幾個N個農場，及M條路來連線這N個農場，及X表示舉辦party的農場求最短路的題，這個題主要是分為去和來兩個過程，這就需要我們用兩次Djikstra演算法，求兩次最小路，然後去加一下求去和的最大的，具體程式碼實現如下，反向回去的路要反向建圖，然後就達到反向的目的

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

const int INF=0x3f3f3f3f;
int p[1005][1005];
int dis1[1005];
int dis2[1005];
int vis[1005];
int main()
{
    int i,j,n,m,x,a,b,c,key,minn;
    scanf("%d%d%d",&n,&m,&x);
    for(i=1;i<=n;i++)
    {
        dis1[i]=INF;
        dis2[i]=INF;
        vis[i]=0;
        for(j=1;j<=n;j++)
            p[i][j]=INF;
    }

    for(i=0;i<m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        p[a][b]=c;
    }
    dis1[x]=0;
    key=x;
    for(i=1;i<=n;i++)
    {
        minn=INF;
        for(j=1;j<=n;j++)
        {
            if(!vis[j]&&dis1[j]<minn)
            {
                minn=dis1[j];
                key=j;
            }
        }
        vis[key]=1;
        for(j=1;j<=n;j++)
        {
            if(minn+p[key][j]<dis1[j])
                dis1[j]=minn+p[key][j];
        }
    }
    for(i=1;i<=n;i++)
    {
        vis[i]=0;
    }
    dis2[x]=0;
    key=x;
    for(i=1;i<=n;i++)
    {
        minn=INF;
        for(j=1;j<=n;j++)
        {
            if(!vis[j]&&dis2[j]<minn)
            {
                minn=dis2[j];
                key=j;
            }
        }
        vis[key]=1;
        for(j=1;j<=n;j++)
        {
            if(minn+p[j][key]<dis2[j])
                dis2[j]=minn+p[j][key];
        }
    }
    minn=0;
    for(i=1;i<=n;i++)
    {
        if(minn<dis1[i]+dis2[i])
            minn=dis1[i]+dis2[i];
    }
    printf("%d\n",minn);
    return 0;
}