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>
#pragma GCC optimize("O3")
using namespace std;
const int MAX = 1e5 + 5;
const long long MAX2 = 11;
const long long MOD = 998244353;
const long long MOD2 = 1000005329;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const long double EPS = 1e-18;
const int block = 2000;
int n, m, p[MAX], a, b, id, ans[MAX], cnt;
pair<pair<int, int>, pair<int, int> > e[MAX];
vector<pair<int, pair<int, int> > > z;
map<pair<int, int>, int> hs;
set<int> sp;
vector<pair<int, int> > v[MAX];
int dfs_num[MAX], dfs_low[MAX], dfs_parent[MAX];
int dfsNumberCounter = 1, dfsRoot, rootChildren;
void articulationPointAndBridge(int u) {
  dfs_low[u] = dfs_num[u] = dfsNumberCounter++;
  for (auto i : v[u]) {
    if (!dfs_num[i.first]) {
      dfs_parent[i.first] = u;
      if (u == dfsRoot) rootChildren++;
      articulationPointAndBridge(i.first);
      if (dfs_low[i.first] > dfs_num[u]) {
        if (!ans[i.second]) ans[i.second] = 2;
      }
      dfs_low[u] = min(dfs_low[u], dfs_low[i.first]);
    } else if (i.first != dfs_parent[u])
      dfs_low[u] = min(dfs_low[u], dfs_num[i.first]);
  }
}
int par(int z) {
  if (p[z] == z) return z;
  return p[z] = par(p[z]);
}
inline void relax() {
  hs.clear();
  sp.clear();
  for (auto i : z) {
    if (hs.count(i.second))
      ans[hs[i.second]] = ans[i.first] = 1;
    else {
      a = par(i.second.second), b = par(i.second.first);
      if (a != b) ++cnt;
      p[a] = b;
      hs[i.second] = i.first;
      v[i.second.first].push_back({i.second.second, i.first});
      v[i.second.second].push_back({i.second.first, i.first});
      sp.insert(i.second.second), sp.insert(i.second.first);
    }
  }
  for (auto i : sp) {
    if (!dfs_num[i]) {
      dfsRoot = i, rootChildren = 0;
      articulationPointAndBridge(i);
    }
  }
  for (auto i : sp) v[i].clear();
  z.clear();
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) p[i] = i;
  for (int i = 1; i <= m; ++i)
    cin >> e[i].second.first >> e[i].second.second >> e[i].first.first,
        e[i].first.second = i;
  sort(e + 1, e + 1 + m);
  for (int i = 1; i <= m; ++i) {
    if (e[i].first.first != e[i - 1].first.first) relax();
    a = par(e[i].second.first), b = par(e[i].second.second),
    id = e[i].first.second;
    if (a == b) {
      ans[id] = -1;
      continue;
    }
    dfs_num[a] = dfs_num[b] = 0;
    if (a > b) swap(a, b);
    z.push_back({id, {a, b}});
  }
  if (!z.empty()) relax();
  assert(cnt == n - 1);
  for (int i = 1; i <= m; ++i) {
    if (ans[i] == -1)
      cout << "none\n";
    else if (ans[i] == 2)
      cout << "any\n";
    else
      cout << "at least one\n";
  }
  return 0;
}
