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;
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O0")
#pragma GCC optimize("O1")
long long pw(int a, int b) {
  long long ret = 1;
  long long mul = a;
  while (b > 0) {
    if (b & 1) ret *= mul;
    mul *= mul;
    b /= 2;
  }
  return ret;
}
long long to_int(string s) {
  long long ret = 0;
  for (int(i) = (0); (i) < (s.size()); (i)++) {
    ret += pw(10, s.size() - i - 1) * (long long)(s[i] - '0');
  }
  return ret;
}
const int MAXN = 1e5 + 15, MAXW = 1e6 + 16, inf = 2e9;
struct ed {
  int s, e, x, ans = 0;
};
int n, m, par[MAXN], rnk[MAXN], tin[MAXN], tme, low[MAXN];
ed e[MAXN];
vector<int> we[MAXW];
vector<pair<int, int> > adj[MAXN];
int get(int v) { return par[v] = (par[v] == v ? v : get(par[v])); }
void uni(int a, int b) {
  a = get(a), b = get(b);
  if (a == b) return;
  if (rnk[a] < rnk[b]) swap(a, b);
  par[b] = a, rnk[a] += (rnk[a] == rnk[b]);
}
void dfs(int v, int id = -1) {
  tin[v] = tme++;
  for (int(i) = (0); (i) < (adj[v].size()); (i)++) {
    int u = adj[v][i].first;
    if (tin[u] != inf and adj[v][i].second != id)
      e[adj[v][i].second].ans = 1, low[v] = min(low[v], tin[u]);
    else if (tin[u] == inf)
      dfs(u, adj[v][i].second), low[v] = min(low[v], low[u]);
  }
  if (id != -1) {
    if (low[v] < tin[v])
      e[id].ans = 1;
    else
      e[id].ans = 2;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> n >> m;
  for (int(i) = (0); (i) < (n); (i)++) par[i] = i, tin[i] = low[i] = inf;
  for (int(i) = (0); (i) < (m); (i)++) {
    int w;
    cin >> e[i].s >> e[i].e >> w;
    e[i].s--, e[i].e--;
    e[i].x = e[i].s ^ e[i].e;
    we[w].push_back(i);
  }
  for (int(i) = (0); (i) < (MAXW); (i)++) {
    bool fl = 0;
    tme = 0;
    for (int(j) = (0); (j) < (we[i].size()); (j)++) {
      int id = we[i][j];
      if (get(e[id].s) != get(e[id].e))
        fl = 1, adj[get(e[id].s)].push_back({get(e[id].e), id}),
        adj[get(e[id].e)].push_back({get(e[id].s), id});
    }
    if (!fl) continue;
    for (int(j) = (0); (j) < (we[i].size()); (j)++) {
      int id = we[i][j];
      if (tin[get(e[id].s)] == inf) dfs(get(e[id].s));
    }
    for (int(j) = (0); (j) < (we[i].size()); (j)++) {
      int id = we[i][j];
      tin[get(e[id].s)] = tin[get(e[id].e)] = low[get(e[id].s)] =
          low[get(e[id].e)] = inf;
      adj[get(e[id].s)].clear(), adj[get(e[id].e)].clear();
      uni(e[id].s, e[id].e);
    }
  }
  for (int(i) = (0); (i) < (m); (i)++) {
    if (e[i].ans == 0)
      cout << "none"
           << "\n";
    else if (e[i].ans == 1)
      cout << "at least one"
           << "\n";
    else
      cout << "any"
           << "\n";
  }
}
