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 n, m, num, Father[100000 + 5], res[100000 + 5], b[100000 + 5],
    d[100000 + 5], Trace[100000 + 5];
pair<int, int> e[100000 + 5];
vector<int> t[1000000 + 5], tmp, c[100000 + 5];
map<pair<int, int>, vector<int> > my_map;
void Init();
void Process();
int main() {
  Init();
  Process();
  return 0;
}
void Init() {
  int p, q, r;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &p, &q, &r);
    e[i] = pair<int, int>(p, q);
    t[r].push_back(i);
  }
}
int Get(int x) {
  if (x != Father[x]) Father[x] = Get(Father[x]);
  return Father[x];
}
void Union(int x, int y) {
  x = Get(x);
  y = Get(y);
  if (x == y) return;
  Father[x] = y;
}
void Add(int p, int q) { c[p].push_back(q); }
void DFS(int i) {
  vector<int>::iterator it;
  b[i] = ++num;
  d[i] = b[i];
  for (it = c[i].begin(); it != c[i].end(); ++it)
    if (*it != Trace[i])
      if (b[*it] == 0) {
        Trace[*it] = i;
        DFS(*it);
        d[i] = min(d[i], d[*it]);
      } else
        d[i] = min(d[i], b[*it]);
}
void Process() {
  int x, y;
  vector<int>::iterator it, itx;
  map<pair<int, int>, vector<int> >::iterator it2;
  for (int i = 1; i <= n; ++i) Father[i] = i;
  for (int i = 1; i <= 1000000; ++i)
    if (t[i].size() > 0) {
      num = 0;
      my_map.clear();
      tmp.clear();
      for (it = t[i].begin(); it != t[i].end(); ++it) {
        x = Get(e[*it].first);
        y = Get(e[*it].second);
        if (x > y) swap(x, y);
        if (x != y) {
          my_map[pair<int, int>(x, y)].push_back(*it);
          tmp.push_back(x);
          tmp.push_back(y);
        }
      }
      for (it2 = my_map.begin(); it2 != my_map.end(); ++it2) {
        Add(it2->first.first, it2->first.second);
        Add(it2->first.second, it2->first.first);
      }
      for (it2 = my_map.begin(); it2 != my_map.end(); ++it2)
        for (it = (it2->second).begin(); it != (it2->second).end(); ++it)
          res[*it] = 1;
      for (it = tmp.begin(); it != tmp.end(); ++it)
        if (Trace[*it] == 0) {
          Trace[*it] = -1;
          DFS(*it);
        }
      for (it = tmp.begin(); it != tmp.end(); ++it)
        if (Trace[*it] != -1)
          if (d[*it] >= b[*it]) {
            x = min(*it, Trace[*it]);
            y = max(*it, Trace[*it]);
            if (my_map[pair<int, int>(x, y)].size() == 1) {
              itx = my_map[pair<int, int>(x, y)].begin();
              res[*itx] = 2;
            }
            Trace[*it] = 0;
          }
      for (it = t[i].begin(); it != t[i].end(); ++it)
        Union(e[*it].first, e[*it].second);
      for (it = tmp.begin(); it != tmp.end(); ++it) {
        c[*it].clear();
        b[*it] = Trace[*it] = 0;
      }
    }
  for (int i = 1; i <= m; ++i)
    if (res[i] == 0)
      printf("none\n");
    else if (res[i] == 1)
      printf("at least one\n");
    else if (res[i] == 2)
      printf("any\n");
}
