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;
#pragma GCC optimize("Ofast")
struct edge {
  int u, v, i;
  edge() {}
  edge(int u, int v, int i) : u(u), v(v), i(i) {}
};
const int maxn = 1e5 + 6, maxw = 1e6 + 6;
int n, m;
vector<edge> edgl[maxw];
int ans[maxn];
int par[maxn], sez[maxn];
int find(int x) {
  while (x != par[x]) x = par[x];
  return x;
}
void unite(int u, int v) {
  u = find(u);
  v = find(v);
  if (sez[u] > sez[v]) swap(u, v);
  par[u] = v;
  sez[v] += sez[u];
}
void getbridge(map<int, vector<pair<int, int>>>& adjl, set<int>& bridges,
               map<int, vector<int>>& to, map<int, int>& lvl, map<int, int>& mn,
               int now, int prev, int d) {
  lvl[now] = mn[now] = d;
  bool cek = 1;
  for (auto p : adjl[now]) {
    int next = p.first, i = p.second;
    if (next == prev) {
      to[now].push_back(i);
      if (cek) {
        cek = 0;
        continue;
      }
    }
    if (lvl.count(next))
      mn[now] = min(mn[now], lvl[next]);
    else {
      getbridge(adjl, bridges, to, lvl, mn, next, now, d + 1);
      mn[now] = min(mn[now], mn[next]);
    }
  }
  if (mn[now] == lvl[now]) {
    for (int i : to[now]) bridges.insert(i);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edgl[w].push_back(edge(u, v, i));
  }
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sez[i] = 1;
  }
  for (int w = 1; w < maxw; w++) {
    if (edgl[w].empty()) continue;
    map<int, vector<pair<int, int>>> adjl;
    for (edge e : edgl[w]) {
      int u = find(e.u), v = find(e.v);
      if (u != v) {
        adjl[u].push_back({v, e.i});
        adjl[v].push_back({u, e.i});
      }
    }
    map<int, int> lvl, mn;
    set<int> bridges;
    for (auto p : adjl) {
      map<int, vector<int>> to;
      if (!lvl.count(p.first))
        getbridge(adjl, bridges, to, lvl, mn, p.first, 0, 0);
    }
    for (edge e : edgl[w]) {
      if (find(e.u) == find(e.v))
        ans[e.i] = 0;
      else if (bridges.count(e.i))
        ans[e.i] = 2;
      else
        ans[e.i] = 1;
    }
    for (edge e : edgl[w]) {
      if (find(e.u) != find(e.v)) unite(e.u, e.v);
    }
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 0)
      cout << "none\n";
    else if (ans[i] == 1)
      cout << "at least one\n";
    else
      cout << "any\n";
  }
}
