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 N = 1e5 + 7;
int n, m;
int par[N];
vector<pair<int, int> > ad[N];
vector<pair<int, pair<int, int> > > es[N * 10];
vector<int> com;
bool mrk[N];
int res[N];
int findpar(int v) {
  if (par[v] == v) return v;
  return par[v] = findpar(par[v]);
}
int jmp[N];
int h[N];
void dfs(int v, int pe) {
  mrk[v] = 1;
  jmp[v] = h[v];
  for (auto p : ad[v]) {
    int e = p.second;
    int u = p.first;
    if (e == pe) continue;
    if (!mrk[u]) {
      h[u] = h[v] + 1;
      dfs(u, e);
      jmp[v] = min(jmp[v], jmp[u]);
    } else {
      jmp[v] = min(jmp[v], h[u]);
      res[e] = 1;
    }
  }
  if (jmp[v] == h[v])
    res[pe] = 2;
  else
    res[pe] = 1;
}
void merge(int v, int u) {
  v = findpar(v);
  u = findpar(u);
  if (v == u) return;
  par[v] = u;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    es[w].push_back({i, {v, u}});
  }
  for (int i = 1; i <= n; i++) par[i] = i;
  for (int w = 1; w < N * 10; w++) {
    com.clear();
    for (auto p : es[w]) {
      p.second.first = findpar(p.second.first);
      p.second.second = findpar(p.second.second);
      if (p.second.first == p.second.second) {
        res[p.first] = -1;
        continue;
      }
      ad[p.second.first].push_back({p.second.second, p.first});
      ad[p.second.second].push_back({p.second.first, p.first});
      com.push_back(p.second.first);
      com.push_back(p.second.second);
      mrk[p.second.first] = jmp[p.second.first] = h[p.second.first] = 0;
      mrk[p.second.second] = jmp[p.second.second] = h[p.second.second] = 0;
    }
    for (auto v : com) {
      if (!mrk[v]) {
        h[v] = 1;
        dfs(v, 0);
      }
    }
    for (auto p : es[w]) {
      ad[findpar(p.second.first)].clear();
      ad[findpar(p.second.second)].clear();
      merge(p.second.first, p.second.second);
    }
  }
  for (int i = 1; i <= m; i++) {
    if (res[i] == -1) cout << "none" << '\n';
    if (res[i] == 1) cout << "at least one" << '\n';
    if (res[i] == 2) cout << "any" << '\n';
  }
  return 0;
}
