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 maxn = 1000 * 100 + 5;
vector<long long> aj[maxn], ind[maxn], c[maxn], e;
long long h[maxn], pr[maxn], cm[maxn], ans[maxn];
pair<long long, long long> enpoin[maxn], eg[maxn];
void mrg(long long u, long long v) {
  if (u == v) return;
  if ((long long)(c[u]).size() > (long long)(c[v]).size()) swap(u, v);
  for (long long i = (long long)(c[u]).size() - 1; i >= 0; i--) {
    long long x = c[u][i];
    c[v].push_back(x);
    cm[x] = v;
  }
  c[u].clear();
}
long long dfs(long long v) {
  long long cut = h[v];
  long long cn = 0;
  for (long long i = 0; i < (long long)(aj[v]).size(); i++) {
    long long u = aj[v][i];
    if (u == pr[v]) cn++;
  }
  for (long long i = 0; i < (long long)(aj[v]).size(); i++) {
    long long u = aj[v][i];
    if (!h[u]) {
      pr[u] = v;
      h[u] = h[v] + 1;
      long long x = dfs(u);
      if (x > h[v])
        ans[ind[v][i]] = 2;
      else
        (cut) = min((cut), (x));
    } else if (u != pr[v] or cn > 1)
      (cut) = min((cut), (h[u]));
  }
  return cut;
}
void new_graph() {
  for (long long i = 0; i < (long long)(e).size(); i++) {
    long long x = e[i], u = cm[enpoin[x].first], v = cm[enpoin[x].second];
    if (!h[u]) {
      h[u] = 1;
      pr[u] = u;
      dfs(u);
    }
  }
  for (long long i = 0; i < (long long)(e).size(); i++) {
    long long x = e[i], u = cm[enpoin[x].first], v = cm[enpoin[x].second];
    mrg(u, v);
    aj[u].clear();
    aj[v].clear();
    ind[u].clear();
    ind[v].clear();
    h[u] = 0;
    h[v] = 0;
  }
  e.clear();
  return;
}
int32_t main() {
  ios_base::sync_with_stdio;
  cin.tie();
  cout.tie();
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    eg[i] = {w, i};
    enpoin[i] = {u, v};
  }
  sort(eg, eg + m);
  for (long long i = 0; i < n; i++) {
    cm[i] = i;
    c[i].push_back(i);
  }
  long long pt = 0;
  while (pt < m) {
    long long y = eg[pt].first;
    while (eg[pt].first == y) {
      long long x = eg[pt].second;
      long long u = enpoin[x].first, v = enpoin[x].second;
      if (cm[u] != cm[v]) ans[x] = 1;
      e.push_back(x);
      aj[cm[v]].push_back(cm[u]);
      aj[cm[u]].push_back(cm[v]);
      ind[cm[v]].push_back(x);
      ind[cm[u]].push_back(x);
      pt++;
    }
    new_graph();
  }
  for (long long i = 0; i < m; i++) {
    if (ans[i] == 2)
      cout << "any" << '\n';
    else if (ans[i] == 1)
      cout << "at least one" << '\n';
    else
      cout << "none" << '\n';
  }
}
