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 + 1, M = 1e6 + 1;
int n, m, dsu[N], pre[N], low[N];
vector<tuple<int, int, int> > edges[M];
vector<pair<int, int> > g[N];
string res[N];
int root(int v) { return (dsu[v] < 0 ? v : dsu[v] = root(dsu[v])); }
int idx;
void dfs(int u, int p) {
  pre[u] = low[u] = ++idx;
  map<int, int> mp;
  for (auto e : g[u]) {
    int v, i;
    tie(v, i) = e;
    if (mp.count(v)) res[mp[v]] = "at least one";
    mp[v] = i;
    if (v == p) continue;
    if (!pre[v]) {
      dfs(v, u);
      low[u] = min(low[v], low[u]);
      if (low[v] > pre[u]) res[i] = "any";
    } else
      low[u] = min(pre[v], low[u]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  fill(dsu + 1, dsu + n + 1, -1);
  fill(res, res + m, string("at least one"));
  for (int i = 0, u, v, w; i < m; i++) {
    cin >> u >> v >> w;
    edges[w].emplace_back(u, v, i);
  }
  for (int w = 1; w < M; w++) {
    vector<int> ls;
    for (auto e : edges[w]) {
      int u, v, i;
      tie(u, v, i) = e;
      u = root(u), v = root(v);
      if (u == v) {
        res[i] = "none";
        continue;
      }
      g[u].emplace_back(v, i);
      g[v].emplace_back(u, i);
      ls.push_back(u);
      ls.push_back(v);
    }
    for (int v : ls)
      if (!pre[v]) dfs(v, -1);
    for (int v : ls) {
      g[v].clear();
      pre[v] = low[v] = 0;
      idx = 0;
    }
    for (auto e : edges[w]) {
      int u, v, i;
      tie(u, v, i) = e;
      u = root(u), v = root(v);
      if (u == v) continue;
      if (dsu[u] > dsu[v]) swap(u, v);
      dsu[u] += dsu[v];
      dsu[v] = u;
    }
  }
  for (int i = 0; i < m; i++) cout << res[i] << '\n';
}
