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 <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")

#define rep(i,n) for(int i=0;i<n;i++)
typedef pair<int,int> P;
const int MOD = 1000000007;

int V,E; //グラフサイズ
int Tmax; //最大時間
int ans[10010] = {}; //出力用
int G[505][505]; //隣接matrix
int pre[505]; //minimum prev node
int disMin[505]; //最小コスト
bool used[505]; //Visited frag


void dijkstra(int start, int goal){
  rep(i, V) disMin[i] = MOD;
  rep(i, V) pre[i] = -1;
  disMin[start] = 0; //初期位置1
  rep(i,V) used[i] = false;

  while( 1 ){
    int v = -1;
    rep(u, V) {
      if( !used[u] && (v==-1 || disMin[u]<disMin[v]) ) v = u;
    }
    if(v == -1) break;
    used[v] = true;

    rep(u, V) {
      if( disMin[u] > disMin[v] + G[v][u] ){
        disMin[u] = disMin[v] + G[v][u];
        pre[u] = v;
      }
    }
  }
  // cout << "   disMin[i]:"; rep(i, V) cout << disMin[i] << " "; cout << "\n";
}
//(求めた最短経路が終了時刻いないなら)、往復の経路復元をする
vector<int> getPath(int g){
  vector<int> path = {};
  for(; g != -1; g = pre[g]) {
    path.push_back(g);
    // cout << g << "\n";
  }
  return path;
}

int main(void){
  cin >> V >> E; //input graph
  rep(i,V) rep(j,V) G[i][j] = MOD;
  rep(i, E){
    int u,v,dis; cin >> u >> v >> dis;
    G[u-1][v-1] = dis; G[v-1][u-1] = dis;
  }
  cin >> Tmax;
  rep(i,Tmax) ans[i] = -1;

  vector<P> order;
  rep(i, Tmax) {
		int orderSum; cin >> orderSum; //その時刻で発生した注文数
		rep(j, orderSum) {
			int new_id, dest; cin >> new_id >> dest; dest -= 1;
      order.push_back({i, dest});
		}
	}
  int i = 0; int j = 0; int now = 0;
  while(i < order.size() && j < Tmax ){
    dijkstra(now, order[i].second);
    // cout << "dis: " << disMin[order[i].second] << "\n";
    // cout << order[i].first << " "<< j <<  "\n";
    if( j + 2*disMin[order[i].second] < Tmax && j >= order[i].first ){
      vector<int> goPath = getPath( order[i].second );
      reverse( goPath.begin(), goPath.end() );
      for(int k = 1; i<goPath.size();k++) {
        rep(l, G[now][goPath[k]]){
          ans[j] = goPath[k]+1;
          j++;
        }
        now = goPath[k];
        i++;
      }
      vector<int> rePath = getPath( now );
      for(int k = 1; i<rePath.size();k++) {
        rep(l, G[now][rePath[k]]){
          ans[j] = rePath[k]+1;
          j++;
        }
        now = rePath[k];
        i++;
      }
    }else if(j + disMin[order[i].second] < Tmax && j >= order[i].first ){
      vector<int> goPath = getPath( order[i].second );
      reverse( goPath.begin(), goPath.end() );
      for(int k = 1; i<goPath.size();k++) {
        rep(l, G[now][goPath[k]]){
          ans[j] = goPath[k]+1;
          j++;
        }
        now = goPath[k+1];
        i++;
      }
      break;
    }else {
      j++;i++;
    }
  }
  rep(i, Tmax)  cout << ans[i] << "\n"; //output
}