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, M = 1e6 + 100, LG = 30, base = 2;
const int MOD = 1e9 + 7, mod = 1e9 + 9;
const long long INF = 1e9, inf = 1e18;
pair<long long, pair<long long, pair<long long, long long> > > edge[N];
vector<pair<long long, long long> > adj[N];
short ans[N];
long long st[N], num[N], par[N], siz[N], k, stt;
bool mark[N], mark3[N];
void Cutedge(int v, int parr, int ind) {
  mark3[v] = 1, st[v] = stt++;
  bool flag = 0;
  int kk = 0;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i].first, indexx = adj[v][i].second;
    if (!mark3[u])
      kk += k, k = 0, Cutedge(u, v, indexx);
    else {
      if ((u == parr && !flag) || st[u] > st[v])
        int temp = 0;
      else
        num[v]++, num[u]--;
      if (u == parr) flag = 1;
    }
  }
  k += num[v] + kk;
  if (k == 0 && ans[ind] != 1 && v != parr) ans[ind] = 3;
  return;
}
int getpar(int v) {
  if (par[v] == -1) return v;
  return par[v] = getpar(par[v]);
}
void mergee(int a, int b) {
  a = getpar(a), b = getpar(b);
  if (a == b) return;
  if (siz[a] > siz[b]) swap(a, b);
  par[a] = b, siz[b] += siz[a];
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int n, m, x, y, w;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> x >> y >> w;
    edge[i] = make_pair(w, make_pair(i, make_pair(--x, --y)));
  }
  sort(edge, edge + m);
  fill(par, par + n, -1);
  fill(siz, siz + n, 1);
  for (int i = 0; i <= m; i++) {
    int u = edge[i].second.second.first, v = edge[i].second.second.second,
        ind = edge[i].second.first;
    if ((i && edge[i].first != edge[i - 1].first) || i == m) {
      for (int j = i - 1; j >= 0 && !mark[j]; j--) {
        int x = getpar(edge[j].second.second.first),
            y = getpar(edge[j].second.second.second);
        adj[x].push_back(make_pair(y, edge[j].second.first));
        adj[y].push_back(make_pair(x, edge[j].second.first));
      }
      for (int j = i - 1; j >= 0 && !mark[j]; j--) {
        if (!mark3[getpar(edge[j].second.second.second)]) {
          int t = getpar(edge[j].second.second.second);
          Cutedge(t, t, 0);
          k = 0;
        }
      }
      for (int j = i - 1; j >= 0 && mark[j] == 0; j--) {
        int uu = getpar(edge[j].second.second.second);
        int vv = getpar(edge[j].second.second.first);
        adj[uu].clear(), adj[vv].clear();
        mark3[uu] = 0, mark3[vv] = 0;
        st[uu] = 0, st[vv] = 0;
        num[uu] = 0, num[vv] = 0;
      }
      for (int j = i - 1; j >= 0 && !mark[j]; j--)
        mark[j] = 1,
        mergee(edge[j].second.second.second, edge[j].second.second.first);
    }
    if (i < m)
      if (getpar(u) == getpar(v))
        ans[ind] = 1;
      else
        ans[ind] = 2;
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 1)
      cout << "none\n";
    else if (ans[i] == 2)
      cout << "at least one\n";
    else
      cout << "any\n";
  }
  return 0;
}
