Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible! 

* Lines 1..1+M: Same format as "Navigation Nightmare"

* Line 2+M: A single integer, K. 1 <= K <= 10,000

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance. 

7 6

1 6 13 E

6 3 9 E

3 5 7 S

4 1 3 N

2 4 20 W

4 7 2 S

3

1 6

1 4

2 6

13

3

36

Farms 2 and 6 are 20+3+13=36 apart. 

#include<cstdio>

#include<vector>

using namespace std;

const int maxn=+; //节点数

const int maxm=+; //边数

const int maxq=+; //查询数

int par[maxn];

int find(int x){return (par[x]==x)?x:(par[x]=find(par[x]));}

struct Edge{

    int u,v,w;

    Edge(int u=,int v=,int w=){this->u=u,this->v=v,this->w=w;}

};

vector<Edge> E;

vector<int> Ge[maxn];

void addedge(int u,int v,int w)

{

    E.push_back(Edge(u,v,w));

    Ge[u].push_back(E.size()-);

}

struct Query{

    int u,v;

    int lca;

    Query(int u=,int v=,int lca=){this->u=u,this->v=v,this->lca=lca;}

};

vector<Query> Q;

vector<int> Gq[maxn];

void addquery(int u,int v)

{

    Q.push_back(Query(u,v));

    Gq[u].push_back(Q.size()-);

}

bool vis[maxn];

int dist[maxn];

void LCA(int u,int d)

{

    par[u]=u; //建立以u为代表元素的集合

    vis[u]=;

    dist[u]=d;

    for(int i=;i<Ge[u].size();i++)

    {

        Edge &e=E[Ge[u][i]]; int v=e.v;

        if(!vis[v])

        {

            LCA(v,d+e.w);

            par[v]=u; //将v的集合并入u的集合

        }

    }

    for(int i=;i<Gq[u].size();i++)

    {

        Query &q=Q[Gq[u][i]]; int v=q.v;

        if(vis[v])

        {

            q.lca=find(v);

            Q[Gq[u][i]^].lca=q.lca;

        }

    }

}

int m,n,k;

int main()

{

    scanf("%d%d",&n,&m);

    for(int i=;i<=m;i++)

    {

        int u,v,w; char d[];

        scanf("%d%d%d%s",&u,&v,&w,d);

        addedge(u,v,w);

        addedge(v,u,w);

    }

    scanf("%d",&k);

    for(int i=;i<=k;i++)

    {

        int u,v;

        scanf("%d%d",&u,&v);

        addquery(u,v);

        addquery(v,u);

    }

    LCA(,);

    for(int i=;i<=k;i++)

    {

        printf("%d\n",dist[Q[(i-)*].u]+dist[Q[(i-)*].v]-*dist[Q[(i-)*].lca]);

    }

}