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;
int a[100100];
int b[100100];
int w[100100];
int o[100100];
int p[100100];
int res[100100];
char ans[][20] = {"none", "at least one", "any"};
struct cmp {
  bool operator()(int a, int b) { return w[a] < w[b]; }
};
int up(int x) { return p[x] == x ? x : p[x] = up(p[x]); }
map<int, vector<int> > e, ids;
map<int, int> ti;
set<int> mark;
int tt;
int dfs(int x) {
  if (ti.count(x)) return ti[x];
  int fup = ti[x] = tt++;
  for (int i = 0; i < e[x].size(); i++) {
    int y = e[x][i];
    int ei = ids[x][i];
    if (mark.count(ei)) continue;
    mark.insert(ei);
    int r = dfs(y);
    if (r < fup) fup = r;
    if (r > ti[x]) res[ei] = 2;
  }
  return fup;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i] >> b[i] >> w[i];
    a[i]--;
    b[i]--;
    o[i] = i;
  }
  sort(o, o + m, cmp());
  for (int i = 0; i < n; i++) p[i] = i;
  for (int i = 0; i < m;) {
    int s = i;
    while (i < m && w[o[i]] == w[o[s]]) i++;
    e.clear();
    ids.clear();
    ti.clear();
    for (int j = s; j < i; j++) {
      int x = up(a[o[j]]);
      int y = up(b[o[j]]);
      if (x == y) continue;
      res[o[j]] = 1;
      e[x].push_back(y);
      ids[x].push_back(o[j]);
      e[y].push_back(x);
      ids[y].push_back(o[j]);
    }
    for (map<int, vector<int> >::iterator it = e.begin(); it != e.end(); ++it)
      dfs(it->first);
    for (int j = s; j < i; j++) {
      int x = up(a[o[j]]);
      int y = up(b[o[j]]);
      if (x == y) continue;
      p[x] = y;
    }
  }
  for (int i = 0; i < m; i++) cout << ans[res[i]] << endl;
  return 0;
}
