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;
struct yal {
  int u, v, id;
  yal(int u_, int v_, int id_) : u(u_), v(v_), id(id_) {}
};
int const N = 1e5 + 20, M = 1e5 + 20, W = 1e6 + 20;
int n, m, par[N], h[N];
bool mark[N];
vector<yal> ed[W];
vector<pair<int, int> > adj[N];
string ans[M];
int root(int v) { return par[v] < 0 ? v : par[v] = root(par[v]); }
void merge(int u, int v) {
  u = root(u), v = root(v);
  if (u == v) return;
  if (par[u] > par[v]) swap(u, v);
  par[u] += par[v];
  par[v] = u;
}
int dfs(int v, int id = -1) {
  mark[v] = true;
  int ret = h[v];
  for (auto ed : adj[v]) {
    int u = ed.first, id2 = ed.second;
    if (id == id2) continue;
    if (!mark[u]) {
      h[u] = h[v] + 1;
      int tmp = dfs(u, id2);
      ret = min(ret, tmp);
      if (tmp > h[v])
        ans[id2] = "any";
      else
        ans[id2] = "at least one";
    } else
      ans[id2] = "at least one";
    ret = min(ret, h[u]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    ed[c].push_back(yal(a, b, i));
  }
  memset(par, -1, sizeof par);
  for (int w = 0; w < W; w++) {
    if (ed[w].empty()) continue;
    vector<int> vec;
    for (yal e : ed[w]) {
      int u = root(e.u), v = root(e.v), id = e.id;
      if (u == v)
        ans[id] = "none";
      else {
        vec.push_back(u);
        vec.push_back(v);
        adj[u].push_back({v, id});
        adj[v].push_back({u, id});
      }
    }
    for (int v : vec)
      if (!mark[v]) h[v] = 0, dfs(v);
    for (yal e : ed[w]) merge(e.u, e.v);
    for (int v : vec) mark[v] = false, adj[v].clear();
    vec.clear();
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
}
