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 maxn = 100000 + 100;
const int maxm = 100000 + 100;
int n, m;
pair<pair<int, int>, pair<int, int> > e[maxm];
int par[maxn];
int mark[maxn];
vector<int> v;
vector<int> adj[maxn];
int dep[maxn];
int ans[maxm];
int root(int x) { return par[x] < 0 ? x : par[x] = root(par[x]); }
bool merge(int x, int y) {
  x = root(x), y = root(y);
  if (x == y) return false;
  if (par[x] < par[y]) swap(x, y);
  par[y] += par[x], par[x] = y;
  return true;
}
int dfs(int x, int parEdge, int depth) {
  dep[x] = depth;
  int uplink = depth;
  for (int i = 0, _n = (int)((int((adj[x]).size()))); i < _n; i++) {
    int y =
        root(e[adj[x][i]].second.first) + root(e[adj[x][i]].second.second) - x;
    if (adj[x][i] == parEdge) continue;
    if (dep[y] != -1)
      uplink = min(uplink, dep[y]);
    else {
      int tmp = dfs(y, adj[x][i], depth + 1);
      if (tmp == depth + 1) ans[e[adj[x][i]].first.second]++;
      uplink = min(uplink, tmp);
    }
  }
  return uplink;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0, _n = (int)(m); i < _n; i++) {
    cin >> e[i].second.first >> e[i].second.second >> e[i].first.first;
    e[i].second.first--, e[i].second.second--;
    e[i].first.second = i;
  }
  sort(e, e + m);
  memset(par, -1, sizeof par);
  for (int i = 0, _n = (int)(m); i < _n; i++) {
    int j;
    v.clear();
    for (j = i; j < m && e[j].first.first == e[i].first.first; j++) {
      int x = root(e[j].second.first), y = root(e[j].second.second);
      if (x == y) continue;
      ans[e[j].first.second]++;
      if (mark[x] < i + 1) {
        mark[x] = i + 1;
        v.push_back(x);
        adj[x].clear();
        dep[x] = -1;
      }
      if (mark[y] < i + 1) {
        mark[y] = i + 1;
        v.push_back(y);
        adj[y].clear();
        dep[y] = -1;
      }
      adj[x].push_back(j), adj[y].push_back(j);
    }
    for (int k = 0, _n = (int)((int((v).size()))); k < _n; k++)
      if (dep[v[k]] == -1) dfs(v[k], -1, 0);
    for (j = i; j < m && e[j].first.first == e[i].first.first; j++)
      merge(e[j].second.first, e[j].second.second);
    i = j - 1;
  }
  for (int i = 0, _n = (int)(m); i < _n; i++)
    cout << (ans[i] == 0   ? "none"
             : ans[i] == 1 ? "at least one"
                           : "any")
         << endl;
  {
    int _;
    cin >> _;
    return 0;
  }
}
