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>
const int maxN = (int)1e4 + 7;
const int maxM = (int)1e5 + 7;
const int maxT = (int)1e6 + 7;
using namespace std;
static int n, m, lab[maxM], low[maxM], num[maxM];
vector<pair<int, int> > adj[maxM];
map<pair<int, int>, int> mp;
string ans[maxM];
vector<pair<int, pair<int, int> > > e;
static bool avail[maxM];
int FindSet(int u) { return lab[u] < 0 ? u : lab[u] = FindSet(lab[u]); }
void Union(int r, int s) {
  if (lab[s] < lab[r]) swap(r, s);
  lab[r] += lab[s];
  lab[s] = r;
}
int DFS(int v, int id = 0, int parent = -1) {
  avail[v] = true;
  low[v] = num[v] = id;
  int cnt = 0;
  for (auto u : adj[v])
    if (!avail[u.first]) {
      int k = DFS(u.first, id + 1, v);
      low[v] = min(k, low[v]);
      if (k > num[v]) ans[u.second] = "any";
    } else if (u.first != parent)
      low[v] = min(low[v], num[u.first]);
    else
      cnt++;
  if (cnt > 1) low[v] = min(low[v], num[v] - 1);
  return low[v];
}
void solve() {
  for (int i = 0; i < m; i++) {
    int j = i;
    while (j < m && e[i].first == e[j].first) j++;
    vector<int> vec;
    for (int k = i; k < j; k++) {
      int l = e[k].second.first;
      int r = e[k].second.second;
      if (FindSet(l) == FindSet(r))
        ans[mp[{l, r}]] = "none";
      else {
        int u = FindSet(l);
        int v = FindSet(r);
        if (!avail[u]) adj[u].clear(), vec.push_back(u);
        if (!avail[v]) adj[v].clear(), vec.push_back(v);
        avail[u] = avail[v] = true;
        adj[u].push_back({v, mp[{l, r}]});
        adj[v].push_back({u, mp[{l, r}]});
      }
    }
    for (auto v : vec) avail[v] = false;
    for (auto v : vec)
      if (!avail[v]) DFS(v);
    for (auto v : vec) avail[v] = false;
    for (int k = i; k < j; k++) {
      int l = FindSet(e[k].second.first);
      int r = FindSet(e[k].second.second);
      if (l != r) Union(l, r);
    }
    i = j - 1;
  }
}
void enter() {
  cin >> n >> m;
  int u, v, w;
  for (int i = 0; i < m; i++) {
    cin >> u >> v >> w;
    u--, v--;
    e.push_back({w, {u, v}});
    mp[{u, v}] = mp[{v, u}] = i;
  }
  sort(e.begin(), e.end());
  memset(lab, -1, sizeof(lab));
}
void Print() {
  for (int i = 0; i < m; i++)
    if (ans[i].empty())
      cout << "at least one\n";
    else
      cout << ans[i] << '\n';
}
int main() {
  auto start = chrono::steady_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  enter();
  solve();
  Print();
  auto end = chrono::steady_clock::now();
  cerr << "In milliseconds : "
       << chrono::duration_cast<chrono::milliseconds>(end - start).count();
  cerr << '\n'
       << "In seconds : "
       << chrono::duration_cast<chrono::seconds>(end - start).count() << '\n';
  return 0;
}
