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 maxn4 = 1e4 + 85, maxn5 = 1e5 + 69, maxn6 = 1e6 + 85,
          maxn7 = 1e7 + 69, maxn8 = 1e8 + 85, maxn9 = 1e9 + 69, modi1 = 1e7 + 7,
          modi2 = 1e9 + 7;
const long long int mod1 = 1e7 + 7, mod2 = 1e9 + 7;
int mn = INT_MAX, mx = INT_MIN;
int fir[maxn5], second[maxn5], weight[maxn5], dp[maxn5], h[maxn5], par[maxn5],
    ans[maxn6];
vector<pair<int, int> > g[maxn5];
vector<int> edge[maxn6];
bool mark[maxn5];
int get_par(int v) { return (par[v] < 0 ? v : par[v] = get_par(par[v])); }
void merge(int v, int u) {
  if ((v = get_par(v)) == (u = get_par(u))) return;
  if (par[v] < par[u]) swap(v, u);
  par[u] = v;
}
void dfs(int v, int p = -1, int indx = -1) {
  mark[v] = 1;
  dp[v] = h[v];
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i].first;
    int c = g[v][i].second;
    if (!mark[u]) {
      h[u] = h[v] + 1;
      dfs(u, v, c);
      ((dp[v]) = min((dp[v]), (dp[u])));
    } else if (c != indx)
      ((dp[v]) = min((dp[v]), (h[u])));
  }
  if (p != -1 and dp[v] == h[v]) ans[indx] = 2;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> fir[i] >> second[i] >> weight[i];
    fir[i]--;
    second[i]--;
    weight[i]--;
    edge[weight[i]].push_back(i);
  }
  memset(par, -1, sizeof(par));
  for (int i = 0; i < 1e6 + 69; i++) {
    vector<int> vertex;
    for (int j = 0; j < edge[i].size(); j++) {
      int indx = edge[i][j];
      int v = fir[indx], u = second[indx];
      if (get_par(v) == get_par(u))
        ans[indx] = 1;
      else {
        int vv = get_par(v), uu = get_par(u);
        vertex.push_back(vv);
        vertex.push_back(uu);
        g[vv].push_back(make_pair(uu, indx));
        g[uu].push_back(make_pair(vv, indx));
      }
    }
    for (int j = 0; j < vertex.size(); j++) {
      if (!mark[vertex[j]]) {
        h[vertex[j]] = 0;
        dfs(vertex[j]);
      }
    }
    for (int j = 0; j < vertex.size(); j++)
      mark[vertex[j]] = 0, g[vertex[j]].clear();
    for (int j = 0; j < edge[i].size(); j++) {
      int indx = edge[i][j];
      int v = fir[indx], u = second[indx];
      merge(v, u);
    }
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 2) cout << "any\n";
    if (ans[i] == 1) cout << "none\n";
    if (ans[i] == 0) cout << "at least one\n";
  }
  return 0;
}
