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;
long long parent[100100];
long long sz[100100];
long long ans[100100];
long long parof(long long v) {
  if (v == parent[v]) return v;
  return parent[v] = parof(parent[v]);
}
void con(long long a, long long b) {
  if (parof(a) == parof(b)) return;
  if (sz[parof(a)] < sz[parof(b)]) {
    sz[parof(b)] += sz[parof(a)];
    parent[parof(a)] = parof(b);
  } else {
    sz[parof(a)] += sz[parof(b)];
    parent[parof(b)] = parof(a);
  }
}
void dfs(long long v, long long p,
         map<long long, vector<pair<long long, long long> > > &g,
         map<long long, bool> &vis, map<long long, long long> &low,
         map<long long, long long> &high, long long ti = 0) {
  vis[v] = 1;
  high[v] = low[v] = ++ti;
  for (auto ch : g[v]) {
    if (!vis[ch.first]) {
      dfs(ch.first, ch.second, g, vis, low, high, ti);
      low[v] = min(low[v], low[ch.first]);
      if (low[ch.first] > high[v]) {
        ans[ch.second] = 2;
      }
    } else if (ch.second != p)
      low[v] = min(low[v], high[ch.first]);
  }
}
int main() {
  ios::sync_with_stdio(0);
  long long tc = 1;
  for (int tt = 1; tt < tc + 1; tt++) {
    long long n, m;
    cin >> n >> m;
    for (int i = 1; i < n + 1; i++) {
      parent[i] = i;
      sz[i] = 1;
    }
    set<pair<pair<long long, long long>, pair<long long, long long> > > s;
    for (int i = 1; i < m + 1; i++) {
      long long a, b, w;
      cin >> a >> b >> w;
      s.insert(make_pair(make_pair(w, i), make_pair(min(a, b), max(a, b))));
    }
    while (s.size()) {
      vector<pair<pair<long long, long long>, pair<long long, long long> > > v;
      pair<pair<long long, long long>, pair<long long, long long> > fr =
          *s.begin();
      long long val = fr.first.first;
      while (s.size() && (*s.begin()).first.first == val) {
        v.push_back(*s.begin());
        s.erase(s.begin());
      }
      map<long long, vector<pair<long long, long long> > > g;
      map<long long, bool> vis;
      map<long long, long long> low, high;
      set<long long> vert;
      long long st = 0;
      for (auto vv : v) {
        long long a = vv.second.first;
        long long b = vv.second.second;
        long long ind = vv.first.second;
        if (parof(a) == parof(b))
          ans[ind] = 1;
        else {
          st = parof(a);
          g[parof(a)].push_back(make_pair(parof(b), ind));
          g[parof(b)].push_back(make_pair(parof(a), ind));
          vert.insert(parof(a));
          vert.insert(parof(b));
        }
      }
      if (st != 0) {
        for (auto ss : vert)
          if (!vis[ss]) dfs(ss, -1, g, vis, low, high);
      }
      for (auto vv : v) {
        long long a = vv.second.first;
        long long b = vv.second.second;
        con(a, b);
      }
    }
    for (int i = 1; i < m + 1; i++)
      if (ans[i] == 1)
        cout << "none" << endl;
      else if (ans[i] == 2)
        cout << "any" << endl;
      else
        cout << "at least one" << endl;
  }
  return 0;
}
