Problem: Requirements

* All inputs are of non-negative integer value.
* T_{\text{max}} = 10000
* 200 \leq |V| \leq 400
* 1.5 |V| \leq |E| \leq 2 |V|
* 1 \leq u_{i}, v_{i} \leq |V| (1 \leq i \leq |E|)
* 1 \leq d_{u_i, v_i} \leq \lceil 4\sqrt{2|V|} \rceil (1 \leq i \leq |E|)
* The given graph has no self-loops / multiple edges and is guaranteed to be connected.
* 0 \leq N_{\text{new}} \leq 1
* 1 \leq \mathrm{new\\_id}_{i} \leq T_{\text{last}}+1 (1 \leq i \leq N_{\text{new}}). Note: If all orders are generated by the order generation rule as explained above, the total number of orders is at most T_{\text{last}}+1. Therefore, the possible range of \mathrm{new\\_id}_{i} should be from 1 to T_{\text{last}}+1.
* 2 \leq \mathrm{dst}_i \leq |V| (1 \leq i \leq N_{\text{new}})
* The order IDs \mathrm{new\\_{id}}_i are unique.

Input Format

Input is provided in the following form:


|V| |E|
u_{1} v_{1} d_{u_{1}, v_{1}}
u_{2} v_{2} d_{u_{2}, v_{2}}
:
u_{|E|} v_{|E|} d_{u_{|E|}, v_{|E|}}
T_{\text{max}}
\mathrm{info}_{0}
\mathrm{info}_{1}
:
\mathrm{info}_{T_{\text{max}}-1}


* In the first line |V| denotes the number of vertices, while |E| denotes the number of edges.
* The following |E| lines denote the edges of the graph. In particular, in the ith line u_{i} and v_{i} denote the edge connecting u_{i} and v_{i} and d_{u_{i}, v_{i}} the corresponding distance.
* The following line denotes the number of steps T_{\text{max}}.



In the following line, \mathrm{info}_t is information about the order from the customer that occurs at time t. \mathrm{info}_t is given in the form:


N_{\text{new}}
\mathrm{new\_id}_1 \mathrm{dst}_1
\mathrm{new\_id}_2 \mathrm{dst}_2
\vdots
\mathrm{new\_id}_{N_{\text{new}}} \mathrm{dst}_{N_{\text{new}}}


* N_{\text{new}} represents the number of new orders which appear at time t.
* The next N_{\text{new}} lines give the newly generated order information. The i-th order information indicates that the order ID \mathrm{new\\_{id}}_i of the new order, while \mathrm{dst}_i denotes the vertex to which the customer wishes the order to be delivered.
* Note: If N_{\text{new}}=0, there are no new lines.

Output Format

The Output expects T_{\text{max}} integers in the format specified below.


\mathrm{command}_{0}
\mathrm{command}_{1}
:
\mathrm{command}_{T_{\text{max}}-1}


In particular, \mathrm{command}_{i} shall specify the movement of the delivery car by using one of the following two options:

1) `stay`, if the car shall not move:


-1


2) `move w`, if the car shall be moved one step towards the neighboring vertex w


w


Note that in case of 2) w has to satisfy the following conditions:

* w \in V
* If the car is at vertex u: \left\\{ u, w \right\\} \in E .
* If the car is on the edge \left\\{ u, v \right\\}, w must either satisfy u = w or v = w.



* * *

Examples

Input




Output




Input

5 7
1 2 5
5 3 4
2 4 8
1 5 1
2 3 3
4 5 3
4 3 9
4
1
1 2
1
2 5
1
3 4
0


Output

2
-1
1
5

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef pair<int, int> P;

const int INF = 100000000;

int n;

int dis[500][500];
int pre[500][500];  //iからjへ行く最短ルートでjの1個前に通るところ
int der[500];

int go(int k, int l){  //kからl
    int res = 0;
    int now = k;
    while(pre[l][now] != -1){
        res += dis[now][pre[l][now]];
        for(int i = 0; i < dis[now][pre[l][now]]; i++) cout << pre[l][now] + 1 << endl;
        now = pre[l][now];
    }
    return res;
}

int roop(int restt){
    vector<int> r;
    r.push_back(0);
    r.push_back(0);
    for(int i = 1; i < n; i++){
        if(der[i] > 0){
            int ins = -1;
            int les = 100000000;
            for(int j = 0; j < (int)r.size() - 1; j++){
                if(dis[r[j]][i] + dis[i][r[j + 1]] < les){
                    ins = j + 1;
                    les = dis[r[j]][i] + dis[i][r[j + 1]];
                }
            }
            r.insert(r.begin() + ins, i);
        }
    }
    if((int)r.size() == 2){
        cout << -1 << endl;
        return 1;
    }
    int res = 0;
    for(int i = 0; i < (int)r.size() - 1; i++){
        res += dis[r[i]][r[i + 1]];
    }
    if(res <= restt){
        fill(der, der + n, 0);
        for(int i = 0; i < (int)r.size() - 1; i++) go(r[i], r[i + 1]);
        return res;
    }
    else{
        int nowp = 0;
        int rt = 0;
        while(true){
            int near = INF;
            int argnear = -1;
            for(int i = 0; i < n; i++){
                if(der[i] != 0 && dis[nowp][i] < near){
                    near = dis[nowp][i];
                    argnear = i;
                }
            }
            if(argnear == -1){
                for(int i = 0; i < restt - rt; i++) cout << -1 << endl;
                return restt;
            }
            else{
                int s = dis[nowp][argnear];
                if(s > restt - rt){
                    for(int i = 0; i < restt - rt; i++) cout << -1 << endl;
                    return restt;
                }
                rt += go(nowp, argnear);
                der[argnear] = 0;
                nowp = argnear;
            }
        }
    }
}

int main()
{
    int m;
    cin >> n >> m;
    int u[1000], v[1000], d[1000];
    vector<P> g[500];
    for(int i = 0; i < m; i++){
        cin >> u[i] >> v[i] >> d[i];
        u[i]--;
        v[i]--;
        g[u[i]].push_back(P(v[i], d[i]));
        g[v[i]].push_back(P(u[i], d[i]));
    }
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++) dis[i][j] = INF;
    }
    for(int i = 0; i < n; i++){
        priority_queue<P, vector<P>, greater<P>> que;
        dis[i][i] = 0;
        pre[i][i] = -1;
        que.push(P(0, i));
        while(que.size()){
            P nowp = que.top();
            que.pop();
            int now = nowp.second;
            if(dis[i][now] < nowp.first) continue;
            for(P nextp : g[now]){
                int next = nextp.first;
                int cost = nextp.second;
                if(dis[i][next] > dis[i][now] + cost){
                    dis[i][next] = dis[i][now] + cost;
                    pre[i][next] = now;
                    que.push(P(dis[i][next], next));
                }
            }
        }
    }
    int t;
    cin >> t;
    int b[10002];
    fill(b, b + t + 1, -1);
    for(int i = 0; i < t; i++){
        int nw;
        cin >> nw;
        if(nw == 1){
            int id, dst;
            cin >> id >> dst;
            b[i] = dst - 1;
        }
    }
    int lastt = -1;
    int nowt = 0;
    while(nowt < t){
        for(int i = lastt + 1; i <= nowt; i++) if(b[i] != -1) der[b[i]]++;
        lastt = nowt;
        nowt += roop(t - nowt);
    }
}