Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<cstdio>#include<vector>#include<queue>#include<functional>using namespace std;#define maxr 100001#define maxn 5002#define inf 100000000int n, r;    //intersections,roadsstruct edge{     int to, cost;     edge(const int a, const int b) :to(a), cost(b){}};vector<edge> g[maxn];    //邻接矩阵typedef pair<int, int> p;priority_queue<p, vector<p>, greater<p>> que;    //优先队列 dijkstraint dis[maxn], dis2[maxn];void solve(){    //initialization    fill(dis, dis + maxn, inf);    fill(dis2, dis2 + maxn, inf);    //    cin >> n >> r;    int a, b, c;    while (r--){        scanf("%d %d %d", &a, &b, &c);        g[a].push_back(edge(b, c));　　//刚开始我只是申请的edge临时变量，到后面调用的时候就会出错，所以直接添加了构造函数        g[b].push_back(edge(a, c));    }    //从 N=1 开始的最短路    dis[1] = 0;    que.push(p(0, 1));    while (!que.empty()){        p pir = que.top(); que.pop();        int v = pir.second, d = pir.first;        if (dis2[v] < d)continue;        for (int i = 0; i < g[v].size(); i++){            edge &e = g[v][i];            int d2 = e.cost + d;            if (dis[e.to]>d2){                swap(dis[e.to], d2);                que.push(p(dis[e.to], e.to));            }            if (dis2[e.to]>d2&&dis[e.to] < d2){                dis2[e.to] = d2;                que.push(p(dis2[e.to], e.to));            }        }    }    printf("%d\n", dis2[n]);}int main(){    solve();//    getchar();    return 0;}