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;
const long long N = 1e5 + 10;
long long n, m, par[N], mark[N], mn[N], depth[N], ans[N];
vector<pair<pair<long long, long long>, pair<long long, long long> > > edge;
vector<pair<long long, long long> > adj[N];
long long get_par(long long v) {
  return (par[v] == v ? v : par[v] = get_par(par[v]));
}
void merge(long long v, long long u) { par[get_par(v)] = get_par(u); }
void dfs(long long v, long long h, long long par_indx) {
  depth[v] = mn[v] = h;
  mark[v] = 1;
  for (pair<long long, long long> i : adj[v]) {
    long long u = i.first, indx = i.second;
    if (indx == par_indx) continue;
    if (mark[u]) {
      if (depth[u] > depth[v]) {
        if (mn[u] <= h)
          ans[indx] = 2;
        else
          ans[indx] = 1;
      }
      mn[v] = min(mn[v], depth[u]);
      continue;
    }
    dfs(u, h + 1, indx);
    mn[v] = min(mn[v], mn[u]);
    if (mn[u] <= h)
      ans[indx] = 2;
    else
      ans[indx] = 1;
  }
}
void solve(long long l, long long r) {
  set<long long> nodes;
  for (long long i = l; i < r; i++) {
    long long v = edge[i].first.first, u = edge[i].first.second;
    v = get_par(v), u = get_par(u);
    edge[i].first = {v, u};
    if (v == u) continue;
    nodes.insert(v);
    nodes.insert(u);
  }
  for (long long i : nodes) {
    adj[i].clear();
    mark[i] = 0;
    par[i] = i;
  }
  for (long long i = l; i < r; i++) {
    long long v = edge[i].first.first, u = edge[i].first.second,
              indx = edge[i].second.second;
    if (v == u) continue;
    adj[v].push_back({u, indx});
    adj[u].push_back({v, indx});
  }
  for (long long i : nodes)
    if (!mark[i]) dfs(i, 0, -1);
  for (long long i = l; i < r; i++)
    merge(edge[i].first.first, edge[i].first.second);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long v, u, c;
    cin >> v >> u >> c;
    v--;
    u--;
    edge.push_back({{v, u}, {c, i}});
  }
  sort(edge.begin(), edge.end(),
       [&](pair<pair<long long, long long>, pair<long long, long long> > p1,
           pair<pair<long long, long long>, pair<long long, long long> > p2) {
         return p1.second.first < p2.second.first;
       });
  for (long long i = 0; i < n; i++) par[i] = i;
  for (long long i = 0; i < m; i++) {
    long long r = i + 1;
    while (r < m && edge[r].second.first == edge[i].second.first) r++;
    solve(i, r);
    i = r - 1;
  }
  for (long long i = 0; i < m; i++)
    if (!ans[i])
      cout << "none\n";
    else if (ans[i] == 1)
      cout << "any\n";
    else
      cout << "at least one\n";
}
