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;
struct edge {
  int v, u, ind;
  edge() {}
  edge(int a, int b, int d) : v(a), u(b), ind(d) {}
};
const int N = 100 * 1000 + 10, A = 1000 * 1000 + 10;
vector<edge> e[A];
vector<pair<int, int> > graph[N];
int par[N], low[N], h[N];
string ans[N];
bool mark[N];
int get_par(int v) {
  if (par[v] == v) return v;
  return par[v] = get_par(par[v]);
}
void dfs(int v, int e) {
  mark[v] = true;
  for (int i = 0; i < (int)graph[v].size(); i++) {
    int u = graph[v][i].first, id = graph[v][i].second;
    if (!mark[u]) {
      h[u] = h[v] + 1;
      dfs(u, id);
      if (low[u] != -1)
        if ((low[v] == -1 && h[low[u]] <= h[v]) || h[low[v]] > h[low[u]])
          low[v] = low[u];
      if (low[u] == -1 || h[low[u]] > h[v]) ans[id] = "any";
    } else if (id != e)
      if (low[v] == -1 || h[u] < h[low[v]]) low[v] = u;
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    e[c].push_back(edge(a, b, i));
    ans[i] = "at least one";
  }
  for (int i = 0; i < n; i++) {
    par[i] = i;
    low[i] = -1;
  }
  for (int i = 0; i < A - 5; i++) {
    vector<int> vec;
    for (int j = 0; j < (int)e[i].size(); j++) {
      int v = get_par(e[i][j].v), u = get_par(e[i][j].u);
      if (v == u) {
        ans[e[i][j].ind] = "none";
        continue;
      }
      graph[v].push_back(make_pair(u, e[i][j].ind));
      graph[u].push_back(make_pair(v, e[i][j].ind));
      vec.push_back(v);
      vec.push_back(u);
    }
    for (int j = 0; j < (int)vec.size(); j++)
      if (!mark[vec[j]]) {
        h[vec[j]] = 0;
        dfs(vec[j], -1);
      }
    for (int j = 0; j < (int)vec.size(); j++) {
      mark[vec[j]] = false;
      graph[vec[j]].clear();
      low[vec[j]] = -1;
    }
    for (int j = 0; j < (int)e[i].size(); j++) {
      int v = get_par(e[i][j].v), u = get_par(e[i][j].u);
      par[u] = v;
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}
