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 long long inf = 5e18 + 7;
const int M = 1e9 + 7;
const int N = 1e6 + 7;
int n, m, v, u, w, p[N], r[N], up[N], tin[N], t;
vector<pair<pair<int, int>, pair<int, int> > > e;
vector<pair<int, int> > g[N];
string ans[N];
int get(int v) {
  if (p[v] == v) return v;
  return p[v] = get(p[v]);
}
void unite(int x, int y) {
  x = get(x);
  y = get(y);
  if (x == y) return;
  if (r[x] < r[y]) swap(x, y);
  p[y] = x;
  r[x] += r[y];
  return;
}
void dfs(int v, int p) {
  up[v] = tin[v] = ++t;
  for (auto to : g[v]) {
    if (to.second == p) continue;
    if (tin[to.first]) {
      up[v] = min(up[v], tin[to.first]);
      ans[to.second] = "at least one";
    } else {
      dfs(to.first, to.second);
      up[v] = min(up[v], up[to.first]);
      if (up[to.first] > tin[v])
        ans[to.second] = "any";
      else
        ans[to.second] = "at least one";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    p[i] = i;
    r[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    cin >> v >> u >> w;
    v--;
    u--;
    e.push_back({{w, i}, {v, u}});
  }
  sort(e.begin(), e.end());
  for (int l = 0; l < m;) {
    int r = l + 1;
    while (r < m && e[r].first.first == e[l].first.first) r++;
    for (int i = l; i < r; i++) {
      int v = e[i].second.first;
      int u = e[i].second.second;
      int nom = e[i].first.second;
      v = get(v);
      u = get(u);
      if (v == u) {
        ans[nom] = "none";
        continue;
      }
      g[v].push_back({u, nom});
      g[u].push_back({v, nom});
    }
    t = 0;
    for (int i = l; i < r; i++) {
      int v = e[i].second.first;
      int u = e[i].second.second;
      v = get(v);
      u = get(u);
      if (v == u) continue;
      if (!tin[v]) dfs(v, -1);
    }
    for (int i = l; i < r; i++) {
      v = e[i].second.first;
      u = e[i].second.second;
      v = get(v);
      u = get(u);
      unite(v, u);
      tin[v] = 0;
      tin[u] = 0;
      g[v].clear();
      g[u].clear();
    }
    l = r;
  }
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
  return 0;
}
