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;
template <typename C>
auto test(C* x) -> decltype(cerr << *x, 0LL);
template <typename C>
char test(...);
template <typename C>
struct itr {
  C begin, end;
};
template <typename C>
itr<C> get_range(C b, C e) {
  return itr<C>{b, e};
}
struct debug {
  template <typename T>
  debug& operator<<(const T&) {
    return *this;
  }
};
string _ARR_(int* arr, int sz) {
  string ret = "{ " + to_string(arr[0]);
  for (int i = 1; i < sz; i++) ret += " , " + to_string(arr[i]);
  ret += " }";
  return ret;
}
const int INF = 1e9 + 7, N = 1e5, M = 1e6;
struct edge {
  int u, v, w, id;
  bool operator<(const edge& oth) const { return w > oth.w; }
};
int n, m, type[N], p[N], ds[N], df[N], dt = 0;
array<int, 3> pr[N];
vector<array<int, 3>> adj[N];
priority_queue<edge> pq;
vector<edge> rem[M];
vector<string> ans = {"none", "at least one", "any"};
int find(int u) { return (p[u] == u ? u : p[u] = find(p[u])); }
bool unite(int v, int u) {
  u = find(u);
  v = find(v);
  if (u == v) return 0;
  p[u] = v;
  return 1;
}
void dfs(int u, int par = 0) {
  ds[u] = dt++;
  for (auto x : adj[u]) {
    int v = x[0], w = x[1], id = x[2];
    if (v != par) {
      pr[v] = {u, w, id};
      dfs(v, u);
    }
  }
  df[u] = dt++;
}
bool in_sub(int u, int v) { return (ds[v] >= ds[u] && df[v] <= df[u]); }
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  iota(p, p + N, 0);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u;
    --v;
    pq.push({u, v, w, i});
  }
  int cnt = 0;
  while (cnt < n - 1) {
    auto e = pq.top();
    pq.pop();
    if (unite(e.u, e.v)) {
      type[e.id] = 2;
      adj[e.u].push_back({e.v, e.w, e.id});
      adj[e.v].push_back({e.u, e.w, e.id});
      ++cnt;
    } else {
      rem[e.w].push_back(e);
    }
  }
  while (pq.size()) {
    auto e = pq.top();
    pq.pop();
    rem[e.w].push_back(e);
  }
  dfs(0);
  for (int i = 1; i < n; ++i) {
    int u = pr[i][0], w = pr[i][1], v = i, id = pr[i][2];
    for (auto e : rem[w]) {
      if (in_sub(v, e.u) ^ in_sub(v, e.v)) {
        type[e.id] = 1;
        type[id] = 1;
      }
    }
  }
  for (int i = 0; i < m; ++i) cout << ans[type[i]] << '\n';
  return 0;
}
