You are given a connected weighted undirected graph without any loops and multiple edges. 

Let us remind you that a graph's spanning tree is defined as an acyclic connected subgraph of the given graph that includes all of the graph's vertexes. The weight of a tree is defined as the sum of weights of the edges that the given tree contains. The minimum spanning tree (MST) of a graph is defined as the graph's spanning tree having the minimum possible weight. For any connected graph obviously exists the minimum spanning tree, but in the general case, a graph's minimum spanning tree is not unique.

Your task is to determine the following for each edge of the given graph: whether it is either included in any MST, or included at least in one MST, or not included in any MST.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105, <image>) — the number of the graph's vertexes and edges, correspondingly. Then follow m lines, each of them contains three integers — the description of the graph's edges as "ai bi wi" (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106, ai ≠ bi), where ai and bi are the numbers of vertexes connected by the i-th edge, wi is the edge's weight. It is guaranteed that the graph is connected and doesn't contain loops or multiple edges.

Output

Print m lines — the answers for all edges. If the i-th edge is included in any MST, print "any"; if the i-th edge is included at least in one MST, print "at least one"; if the i-th edge isn't included in any MST, print "none". Print the answers for the edges in the order in which the edges are specified in the input.

Examples

Input

4 5
1 2 101
1 3 100
2 3 2
2 4 2
3 4 1


Output

none
any
at least one
at least one
any


Input

3 3
1 2 1
2 3 1
1 3 2


Output

any
any
none


Input

3 3
1 2 1
2 3 1
1 3 1


Output

at least one
at least one
at least one

Note

In the second sample the MST is unique for the given graph: it contains two first edges.

In the third sample any two edges form the MST for the given graph. That means that each edge is included at least in one MST.
#include <bits/stdc++.h>
using namespace std;
int N, M;
int Gr[100000], Ans[100000];
vector<pair<pair<int, int>, pair<int, int> > > Edges;
int get_par(int x) {
  int head = x, tmp;
  while (Gr[head] != head) head = Gr[head];
  while (x != head) {
    tmp = Gr[x];
    Gr[x] = head;
    x = tmp;
  }
  return head;
}
int main() {
  int i, j, a, b, w;
  cin >> N >> M;
  for (i = 0; i < (int)(N); i++) Gr[i] = i, Ans[i] = 0;
  for (i = 0; i < (int)(M); i++) {
    cin >> a >> b >> w;
    Edges.push_back(make_pair(make_pair(w, i), make_pair(a - 1, b - 1)));
  }
  sort(Edges.begin(), Edges.end());
  vector<int> done = vector<int>(N, -1);
  vector<int> from = vector<int>(N, -1);
  vector<vector<pair<int, int> > > con =
      vector<vector<pair<int, int> > >(N, vector<pair<int, int> >());
  for (i = 0; i < (int)(M); i++) {
    for (j = i; j < M; j++)
      if (Edges[i].first.first != Edges[j].first.first) break;
    int st = i, en = j - 1;
    i = en;
    set<int> cares;
    for (j = st; j <= en; j++) {
      if (get_par(Edges[j].second.first) == get_par(Edges[j].second.second))
        continue;
      done[Gr[Edges[j].second.first]] = from[Gr[Edges[j].second.first]] = -1;
      con[Gr[Edges[j].second.first]].clear();
      cares.insert(Gr[Edges[j].second.first]);
      done[Gr[Edges[j].second.second]] = from[Gr[Edges[j].second.second]] = -1;
      con[Gr[Edges[j].second.second]].clear();
      cares.insert(Gr[Edges[j].second.second]);
    }
    for (j = st; j <= en; j++)
      if (Gr[Edges[j].second.first] != Gr[Edges[j].second.second]) {
        con[Gr[Edges[j].second.first]].push_back(
            make_pair(Gr[Edges[j].second.second], Edges[j].first.second));
        con[Gr[Edges[j].second.second]].push_back(
            make_pair(Gr[Edges[j].second.first], Edges[j].first.second));
        Ans[Edges[j].first.second] = 1;
      }
    for (set<int>::iterator it = cares.begin(); it != cares.end(); ++it) {
      if (done[*it] != -1) continue;
      vector<pair<int, int> > stack, cycle;
      stack.push_back(make_pair(*it, 0));
      done[*it] = ((int)(stack).size());
      while (((int)(stack).size())) {
        int x = stack.back().first;
        int idx = stack.back().second++;
        if (idx == ((int)(con[x]).size())) {
          stack.pop_back();
          if (((int)(cycle).size()) != 0 && cycle.back().second == done[x]) {
            Ans[con[stack.back().first][stack.back().second - 1].second] = 2;
            cycle.back().second--;
            if (cycle.back().first == cycle.back().second) cycle.pop_back();
          }
        } else if (con[x][idx].second != from[x]) {
          if (done[con[x][idx].first] == -1) {
            from[con[x][idx].first] = con[x][idx].second;
            stack.push_back(make_pair(con[x][idx].first, 0));
            done[con[x][idx].first] = ((int)(stack).size());
          } else if (done[con[x][idx].first] < done[x]) {
            Ans[con[x][idx].second] = 2;
            if (((int)(cycle).size()) == 0 ||
                cycle.back().second < done[con[x][idx].first]) {
              cycle.push_back(make_pair(done[con[x][idx].first], done[x]));
            } else {
              cycle.back().first =
                  min(cycle.back().first, done[con[x][idx].first]);
              cycle.back().second = done[x];
            }
          } else {
            Ans[con[x][idx].second] = 2;
          }
        }
      }
    }
    for (j = st; j <= en; j++) {
      if (get_par(Edges[j].second.first) == get_par(Edges[j].second.second))
        continue;
      Gr[Gr[Edges[j].second.first]] = Gr[Edges[j].second.second];
    }
  }
  for (i = 0; i < (int)(M); i++) {
    if (Ans[i] == 0)
      cout << "none" << endl;
    else if (Ans[i] == 1)
      cout << "any" << endl;
    else
      cout << "at least one" << endl;
  }
  return 0;
}
