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 + 100;
int par[N], ans[N], v[N], u[N], w[N], a[N], h[N];
vector<pair<int, int> > adj[N];
vector<int> vec;
bool mark[N];
int n, m;
int getPar(int v) { return (par[v] == v ? v : par[v] = getPar(par[v])); }
void merge(int v, int u) {
  v = getPar(v);
  u = getPar(u);
  par[u] = v;
}
void add(int a, int b, int i) {
  if (a == b) return;
  vec.push_back(a);
  adj[a].push_back({b, i});
}
int dfs(int v, int ind) {
  mark[v] = true;
  int mn = h[v];
  for (auto [u, i] : adj[v]) {
    if (i == ind) continue;
    if (mark[u]) {
      mn = min(mn, h[u]);
      continue;
    }
    h[u] = h[v] + 1;
    int tmp = dfs(u, i);
    if (tmp > h[v]) ans[i] = 2;
    mn = min(mn, tmp);
  }
  return mn;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> v[i] >> u[i] >> w[i];
    v[i]--;
    u[i]--;
    a[i] = i;
  }
  sort(a, a + m, [&](int i, int j) { return w[i] < w[j]; });
  for (int i = 0; i < n; i++) par[i] = i;
  for (int i = 0; i < m;) {
    int j = i;
    for (; j < m && w[a[j]] == w[a[i]]; j++) {
      v[a[j]] = getPar(v[a[j]]);
      u[a[j]] = getPar(u[a[j]]);
      add(v[a[j]], u[a[j]], a[j]), add(u[a[j]], v[a[j]], a[j]);
    }
    for (; i < j; i++) merge(v[a[i]], u[a[i]]);
    for (auto v : vec)
      if (!mark[v]) dfs(v, m);
    for (auto v : vec) {
      for (auto [u, i] : adj[v])
        if (!ans[i]) ans[i] = 1;
      h[v] = 0;
      adj[v].clear();
      mark[v] = false;
      vec.clear();
    }
  }
  for (int i = 0; i < m; i++) {
    if (!ans[i])
      cout << "none\n";
    else
      cout << (--ans[i] ? "any" : "at least one") << '\n';
  }
}
