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 MX = 1e5 + 9;
long long w[MX], ans[MX], par[MX], up[MX], dis[MX], mrk[MX];
pair<long long, long long> edg[MX], a[MX];
vector<pair<long long, long long> > adj[MX];
long long FS(long long v) {
  if (par[v] < 0) return v;
  return par[v] = FS(par[v]);
}
void US(long long a, long long b) {
  a = FS(a), b = FS(b);
  if (a == b) return;
  if (par[a] > par[b]) swap(a, b);
  par[a] += par[b];
  par[b] = a;
}
void DFS(long long u, long long p) {
  mrk[u] = 1;
  up[u] = dis[u];
  for (pair<long long, long long> v : adj[u]) {
    if (!mrk[v.first]) {
      dis[v.first] = dis[u] + 1;
      DFS(v.first, v.second);
      if (up[v.first] <= dis[u])
        ans[v.second] = 1;
      else
        ans[v.second] = 2;
      up[u] = min(up[u], up[v.first]);
    } else if (v.second != p) {
      up[u] = min(up[u], dis[v.first]);
      ans[v.second] = 1;
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> edg[i].first >> edg[i].second >> w[i];
    a[i] = {w[i], i};
  }
  fill(par, par + MX, -1);
  sort(a, a + m);
  long long j = 0;
  for (long long i = 0; i < m; i = j) {
    vector<long long> v;
    set<long long> s;
    while (j < m && a[i].first == a[j].first) {
      long long ind = a[j].second, f = FS(edg[ind].first),
                g = FS(edg[ind].second);
      if (f == g)
        ans[ind] = -1;
      else {
        s.insert(f), s.insert(g);
        v.push_back(ind);
        adj[f].push_back({g, ind}), adj[g].push_back({f, ind});
      }
      j++;
    }
    while (!s.empty()) {
      long long u = *s.begin();
      s.erase(s.begin());
      if (!mrk[u]) DFS(u, u);
    }
    for (long long x : v) {
      long long f = FS(edg[x].first), g = FS(edg[x].second);
      up[f] = up[g] = dis[f] = dis[g] = mrk[f] = mrk[g] = 0;
      adj[g].clear(), adj[f].clear();
      US(f, g);
    }
  }
  for (long long i = 0; i < m; i++) {
    if (ans[i] == -1) cout << "none" << '\n';
    if (ans[i] == 1) cout << "at least one" << '\n';
    if (ans[i] == 2) cout << "any" << '\n';
  }
  return 0;
}
