cf 864 F. Cities Excursions

F. Cities Excursions

There are n cities in Berland. Some pairs of them are connected with m directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities (x, y) there is at most one road from x to y.

A path from city s to city t is a sequence of cities p₁, p₂, ... , p_k, where p₁ = s, p_k = t, and there is a road from city p_i to city p_i + 1 for each i from 1 to k - 1. The path can pass multiple times through each city except t. It can't pass through t more than once.

A path p from s to t is ideal if it is the lexicographically minimal such path. In other words, p is ideal path from s to t if for any other path q from s to t p_i < q_i, where i is the minimum integer such that p_i ≠ q_i.

There is a tourist agency in the country that offers q unusual excursions: the j-th excursion starts at city s_j and ends in city t_j.

For each pair s_j, t_j help the agency to study the ideal path from s_j to t_j. Note that it is possible that there is no ideal path from s_j to t_j. This is possible due to two reasons:

• there is no path from s_j to t_j;
• there are paths from s_j to t_j, but for every such path p there is another path q from s_j to t_j, such that p_i > q_i, where i is the minimum integer for which p_i ≠ q_i.

The agency would like to know for the ideal path from s_j to t_j the k_j-th city in that path (on the way from s_j to t_j).

For each triple s_j, t_j, k_j (1 ≤ j ≤ q) find if there is an ideal path from s_j to t_j and print the k_j-th city in that path, if there is any.

Input

The first line contains three integers n, m and q (2 ≤ n ≤ 3000,0 ≤ m ≤ 3000, 1 ≤ q ≤ 4·10⁵) — the number of cities, the number of roads and the number of excursions.

Each of the next m lines contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), denoting that the i-th road goes from city x_i to city y_i. All roads are one-directional. There can't be more than one road in each direction between two cities.

Each of the next q lines contains three integers s_j, t_j and k_j (1 ≤ s_j, t_j ≤ n, s_j ≠ t_j, 1 ≤ k_j ≤ 3000).

Output

In the j-th line print the city that is the k_j-th in the ideal path from s_j to t_j. If there is no ideal path from s_j to t_j, or the integer k_j is greater than the length of this path, print the string '-1' (without quotes) in the j-th line.

Example

Input

7 7 5
1 2
2 3
1 3
3 4
4 5
5 3
4 6
1 4 2
2 6 1
1 7 3
1 3 2
1 3 5

Output

2
-1
-1
2
-1
找字典序最小的路径中，经过的第k个城市，可以采用LCA的处理方式，将查询结果按照分类保存，减少递归次数。题目中可能存在自环。需要特判。Tarjan算法的应用。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
struct point{int s,t,k,id;}q[];
vector<point>fq[];
vector<int>v[];
int dnf[],low[],vis[],pos[],x,y;
int val[],coun,num,n,m,r,k;
bool cmp(point a,point b)
{
    return a.s<b.s;
}
void tarjan(int u,int fa)
{
    dnf[u]=++coun;
    low[u]=INF;
    vis[u]=;
    pos[num++]=u;
    if(fa)
    {
        for(int i=;i<fq[u].size();i++)
            if(fq[u][i].k<=num) val[fq[u][i].id]=pos[fq[u][i].k-];
    }
    for(int i=;i<v[u].size();i++)
    {
        if(!dnf[v[u][i]])
        {
            tarjan(v[u][i],fa && dnf[u]<low[u]);//防止自环
            low[u]=min(low[v[u][i]],low[u]);
        }
        else if(vis[v[u][i]]) low[u]=min(low[u],dnf[v[u][i]]);
    }
    vis[u]=;
    --num;
}
int main()
{
    scanf("%d%d%d",&n,&m,&r);
    memset(val,-,sizeof(val));
    for(int i=;i<m;i++)
    {
        scanf("%d%d",&x,&y);
        v[x].push_back(y);
    }
    for(int i=;i<=n;i++)
    {
        sort(v[i].begin(),v[i].end());
    }
    for(int i=;i<r;i++)
    {
        scanf("%d%d%d",&x,&y,&k);
        q[i]=(point){x,y,k,i};
    }
    sort(q,q+r,cmp);
    for(int i=;i<r;i++)
    {
        fq[q[i].t].push_back(q[i]);
        if(q[i].s!=q[i+].s)
        {
            coun=num=;
            memset(dnf,,sizeof(dnf));
            memset(low,,sizeof(low));
            memset(vis,,sizeof(vis));
            tarjan(q[i].s,);
            for(int j=;j<=n;j++) fq[j].clear();
        }
    }
    for(int i=;i<r;i++)
        printf("%d\n",val[i]);
    return ;
}