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
#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define BIT(X,i) ((X>>i)&1)
#define REP(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define all(a) (a).begin(), (a).end()
#define r_all(a) (a).rbegin(), (a).rend()
#define ina(x,n) rep(x_i,n) cin>>x[x_i]
#define outa(x,n) rep(x_i,n) cerr<<x[x_i]<<" \n"[x_i==n-1]

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll, ll> pll;
typedef vector<pll> vll;
typedef priority_queue<pll> pq;

const int MOD = 1e9+7;
const int INF = (1<<30);
const ll INFL = (1L<<61);
const ll SKIP = 1;

int V, E, T;
vl ids, dsts;
vvl edges;
vector<bool> go;
vl depth;
vl visited_num;
vl ans;

void input();

vl findPath(int dest);

int main() {
  input();

  pq que;
  que.push(mp(0, 0));
  while (!que.empty()) {
    pll x = que.top(); que.pop();
    ll dep = x.ff, node = x.ss;
    if (depth[node] == INF) {
      // cout << node+1 << ": " << -dep << endl;
      depth[node] = -dep;
    }
    rep(i,V) {
      if (edges[node][i] != -1 && depth[i] == INF) {
        que.push(mp(dep - 1, i));
      }
    }
  }

  int t = 0, dest;
  vl pri;
int test1 = 0;
  while (t < T) {
    while (t<SKIP || (t<T && pri.empty())) {
      if (dsts[t] != 0) {
        pri.pb(dsts[t]);
        go[dsts[t]] = true;
      }
      ans.pb(-1);
      t++;
    }

    dest = pri[0];
    vl path1 = findPath(dest);
    vl path2 = findPath(dest);
if (test1 != 0) {
  cerr << "t: " << t << endl;
  cerr << "path1: "; outa(path1, path1.size());
  cerr << "path2: "; outa(path2, path2.size());
  cerr << "pri : "; outa(pri, pri.size());
  cerr << "visited_num : "; outa(visited_num, visited_num.size());
  test1--;
}
    reverse(all(path1));

    int node = 0, nt = t;
    REP(i,1,path1.size()) {
      int next_node = path1[i];
      rep(j,edges[node][next_node]) ans.pb(next_node+1);
      nt += edges[node][next_node];
      node = next_node;
    }
    node = dest;
    REP(i,1,path2.size()) {
      int next_node = path2[i];
      rep(j,edges[node][next_node]) ans.pb(next_node+1);
      nt += edges[node][next_node];
      node = next_node;
    }

    vl n_pri;
    if (!pri.empty()) {
      rep(i,pri.size()) {
        if (go[pri[i]]) {
          n_pri.pb(pri[i]);
        }
      }
    }
    for (int i = t+1; i <= min(nt,T-1); i++) {
      if (dsts[i] != 0) {
        n_pri.pb(dsts[i]);
        go[dsts[i]] = true;
      }
    }
    pri = n_pri;
    t = nt;
  }

  int len = ans.size();
  if (len < T) {
    rep(i,T-len) cout << -1 << '\n';
    rep(i,len) cout << ans[i] << '\n';
  } else {
    rep(i,T) cout << ans[i] << '\n';
  }
  return 0;
}



// input
void input() {
  cin >> V >> E;
  edges.assign(V, vl(V, -1));
  go.assign(V, false);
  depth.assign(V, INF);
  visited_num.assign(V,0);

  rep(i,E) {
    int u, v, d;
    scanf("%d %d %d", &u, &v, &d);
    edges[u-1][v-1] = edges[v-1][u-1] = d;
  }
  cin >> T;
  ids.resize(T);
  dsts.resize(T);
  rep(i,T) {
    int n; scanf("%d", &n);
    if (n == 0) {
      ids[i] = 0; dsts[i] = 0;
    } else {
      scanf("%d %d", &ids[i], &dsts[i]);
      dsts[i]--;
    }
  }
}

// find a path
vl findPath(int dest) {
  vl path(1, dest);
  int now = dest;
  go[dest] = false;

  while (now != 0) {
    int nnode = 0, mn = INF;
    vl cands;
    for (int next = 0; next < V; next++) {
      if (edges[now][next] != -1 && depth[next] == depth[now] - 1) {
        cands.pb(next);
      }
    }
    rep(i,cands.size()) {
      if (visited_num[cands[i]] < mn) {
        mn = visited_num[cands[i]];
        nnode = cands[i];
      }
      // if (go[cands[i]]) {
      //   nnode = cands[i];
      //   break;
      // }
    }
    path.pb(nnode);
    go[nnode] = false;
    now = nnode;
    visited_num[nnode]++;
  }

  return path;
}