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 N = 2e5 + 5, inf = 1e15, mod = 1e9 + 7;
long long par[N], sz[N], tmp, l, con[N], d[N], h[N], b[N];
long long id = 0;
vector<long long> vec[N];
map<long long, bool> vst;
long long ans[N];
map<pair<long long, long long>, bool> isc;
map<pair<long long, long long>, int> mp;
map<pair<long long, long long>, int> is;
map<pair<long long, long long>, int> idd;
pair<long long, pair<pair<long long, long long>, long long> > p[N];
void dfs(long long v, long long p) {
  d[v] = h[v];
  con[v] = 1;
  vst[v] = 1;
  for (auto e : vec[v]) {
    if (!con[e]) {
      h[e] = h[v] + 1;
      dfs(e, v);
      d[v] = min(d[v], d[e]);
      if (d[e] > h[v]) {
        isc[{v, e}] = 1;
        isc[{e, v}] = 1;
      }
    } else if (e != p) {
      d[v] = min(d[v], h[e]);
    }
  }
  con[v] = 2;
}
long long root(long long x) { return par[x] == x ? x : par[x] = root(par[x]); }
void mrg(long long x, long long y) {
  x = root(x);
  y = root(y);
  if (x != y) {
    if (sz[x] < sz[y]) swap(x, y);
    par[y] = x;
    sz[x] += sz[y];
  }
}
void f(long long i) {
  for (int j = l; j < i; j++) {
    long long u = root(p[j].second.first.first),
              v = root(p[j].second.first.second);
    if (u != v) {
      if (!vst[u]) {
        b[id++] = u;
        vst[u] = 1;
      }
      if (!vst[v]) {
        b[id++] = v;
        vst[v] = 1;
      }
      vec[u].push_back(v);
      vec[v].push_back(u);
      mp[{u, v}]++;
      mp[{v, u}]++;
    } else
      ans[p[j].second.second] = 1;
  }
  vst.clear();
  for (int j = 0; j < id; j++)
    if (!vst[b[j]]) dfs(b[j], 0);
  for (int j = 0; j < id; j++) {
    con[b[j]] = 0;
    vec[b[j]].clear();
  }
  for (pair<pair<long long, long long>, int> e : mp) {
    if (e.second > 1) {
      isc[{e.first.first, e.first.second}] = 0;
      isc[{e.first.second, e.first.first}] = 0;
    }
  }
  for (int j = l; j < i; j++) {
    long long u = root(p[j].second.first.first),
              v = root(p[j].second.first.second);
    if (isc[{u, v}]) ans[p[j].second.second] = 3;
  }
  for (int j = l; j < i; j++) {
    if (ans[j] == 0) ans[j] = 2;
    mrg(p[j].second.first.first, p[j].second.first.second);
  }
  vst.clear();
  id = 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    p[i] = {w, {{u, v}, i}};
    idd[{u, v}] = i;
    idd[{v, u}] = i;
  }
  sort(p, p + m);
  tmp = p[0].first;
  l = 0;
  for (int i = 0; i < m; i++) {
    if (p[i].first != tmp) {
      f(i);
      mp.clear();
      is.clear();
      isc.clear();
      tmp = p[i].first;
      l = i;
      i--;
    }
  }
  f(m);
  string s[4];
  s[1] = "none";
  s[3] = "any";
  s[2] = "at least one";
  for (int i = 0; i < m; i++) cout << s[ans[i]] << endl;
  return 0;
}
