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 int maxn = 1e5 + 10, maxv = 1e6 + 10;
int n, m, par[maxn], type[maxn], dp[maxn], h[maxn];
vector<pair<pair<int, int>, int>> w[maxv];
vector<pair<int, int>> adj[maxn];
vector<int> vec;
int mark[maxn];
string ans[3];
int get_par(int v) { return v == par[v] ? v : par[v] = get_par(par[v]); }
void merge(int v, int u) {
  int gv = get_par(v), gu = get_par(u);
  if (gv != gu) par[gv] = gu;
  return;
}
void dfs(int v, int he, int e = -1) {
  ++mark[v];
  dp[v] = he;
  h[v] = he;
  for (auto u : adj[v])
    if (!mark[u.first]) {
      dfs(u.first, he + 1, u.second);
      if (dp[u.first] <= he) type[u.second] = 1;
      dp[v] = min(dp[v], dp[u.first]);
    } else if (u.second != e) {
      type[u.second] = 1;
      dp[v] = min(dp[v], h[u.first]);
    }
  return;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) par[i] = i;
  for (int i = 0, v, u, we; i < m; ++i) {
    cin >> v >> u >> we;
    w[we].push_back({{v - 1, u - 1}, i});
  }
  for (int i = 1; i < maxv; ++i)
    if (!w[i].empty()) {
      for (auto j : w[i]) {
        int v = j.first.first, u = j.first.second;
        int gv = get_par(v), gu = get_par(u);
        if (gv == gu) {
          type[j.second] = 2;
          continue;
        }
        vec.push_back(gu);
        vec.push_back(gv);
        adj[gv].push_back({gu, j.second});
        adj[gu].push_back({gv, j.second});
      }
      for (auto j : vec)
        if (!mark[j]) dfs(j, 0);
      for (auto j : w[i]) {
        int v = j.first.first, u = j.first.second;
        merge(v, u);
      }
      for (auto j : vec) {
        mark[j] = 0;
        adj[j].clear();
      }
      vec.clear();
    }
  ans[0] = "any";
  ans[1] = "at least one";
  ans[2] = "none";
  for (int i = 0; i < m; ++i) cout << ans[type[i]] << '\n';
  return 0;
}
