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 mod = 1e15;
const long long mx = 2e5;
const long long inf = 2e9 + 1;
const int sz = 26;
int time1 = 0;
vector<bool> good(mx, false), bad(mx, false);
struct DisjointSet {
  vector<int> parent, rank;
  DisjointSet() {
    parent.resize(mx);
    rank.resize(mx);
    for (int i = 0; i < mx; ++i) parent[i] = i, rank[i] = 0;
  }
  int find_r(int i) {
    if (parent[i] != i) {
      parent[i] = find_r(parent[i]);
    }
    return parent[i];
  }
  void Union(int i, int j) {
    int repi = find_r(i), repj = find_r(j);
    if (repi == repj) return;
    if (rank[repi] > rank[repj]) {
      parent[repj] = repi;
    } else if (rank[repj] > rank[repi]) {
      parent[repi] = repj;
    } else {
      parent[repj] = repi;
      rank[repi]++;
    }
  }
} ds;
map<int, vector<pair<int, int> > > adj;
map<int, int> dist, low;
void dfs(int u, int p = -1, int p_ind = -1) {
  dist[u] = low[u] = ++time1;
  for (auto it : adj[u]) {
    int v = it.first;
    int ind = it.second;
    if (v == p && ind == p_ind) continue;
    if (!dist[v]) {
      dfs(v, u, ind);
      if (dist[u] < low[v]) good[ind] = true;
      low[u] = min(low[u], low[v]);
    } else {
      low[u] = min(dist[v], low[u]);
    }
  }
}
struct rep {
  int src, des, ind;
};
void input() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  map<int, vector<rep> > edge;
  for (int i = 0; i < m; ++i) {
    int s, d, w;
    cin >> s >> d >> w;
    --s;
    --d;
    if (s > d) {
      swap(s, d);
    }
    rep r;
    r.src = s;
    r.des = d;
    r.ind = i;
    edge[w].push_back(r);
  }
  for (int i = 0; i < n; ++i) {
    if (!dist[i]) dfs(i);
  }
  for (auto &it : edge) {
    adj.clear();
    low.clear();
    dist.clear();
    time1 = 0;
    for (rep r : it.second) {
      int s = r.src;
      int d = r.des;
      int a = ds.find_r(s);
      int b = ds.find_r(d);
      if (a == b)
        bad[r.ind] = true;
      else {
        adj[a].push_back(make_pair(b, r.ind));
        adj[b].push_back(make_pair(a, r.ind));
      }
    }
    for (auto &it : adj) {
      if (dist.find(it.first) == dist.end()) {
        dfs(it.first);
      }
    }
    for (rep r : it.second) {
      if (ds.find_r(r.src) != ds.find_r(r.des)) ds.Union(r.src, r.des);
    }
  }
  for (int i = 0; i < m; ++i) {
    if (good[i])
      cout << "any"
           << "\n";
    else if (bad[i])
      cout << "none"
           << "\n";
    else
      cout << "at least one"
           << "\n";
  }
}
int main() {
  input();
  return 0;
}
