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 DSU {
  int* pa;
  int N;
  DSU(int n) {
    n += 1000;
    pa = new int[n];
    fill_n(pa, n, -1);
    N = n;
  }
  void reset() { fill_n(pa, N, -1); }
  int root(int i) {
    if (pa[i] < 0) return i;
    return pa[i] = root(pa[i]);
  }
  bool same(int a, int b) { return root(a) == root(b); }
  int size(int i) { return -pa[root(i)]; }
  void merge(int a, int b) {
    a = root(a);
    b = root(b);
    if (a != b) {
      if (pa[a] >= pa[b]) {
        pa[b] += pa[a];
        pa[a] = b;
      } else {
        pa[a] += pa[b];
        pa[b] = a;
      }
    }
  }
};
vector<pair<pair<int, int>, int>> A[1000001];
string ans[101010];
vector<pair<int, int>> adj[101010];
int timer = 2;
int pirmas[101010];
int maxi[101010];
void dfs(int i, int negalima) {
  if (pirmas[i] != 0) return;
  pirmas[i] = maxi[i] = timer++;
  for (pair<int, int> j : adj[i]) {
    if (j.second == negalima) continue;
    dfs(j.first, j.second);
    if (pirmas[j.first] > pirmas[i]) {
      maxi[i] = min(maxi[i], maxi[j.first]);
      if (maxi[j.first] > pirmas[i]) {
        ans[j.second] = "any";
      }
    } else {
      maxi[i] = min(maxi[i], pirmas[j.first]);
    }
  }
}
int main() {
  clock_t pradzia = clock();
  srand(pradzia);
  int n, m;
  cin >> n >> m;
  int tim = 0;
  for (int i = 0; i < m; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    A[w].push_back({{a, b}, tim++});
  }
  fill_n(ans, m, "at least one");
  DSU X(n + 1);
  for (int w = 1; w <= 1000000; w++) {
    for (pair<pair<int, int>, int> i : A[w]) {
      if (X.same(i.first.first, i.first.second))
        ans[i.second] = "none";
      else {
        adj[X.root(i.first.first)].push_back(
            {X.root(i.first.second), i.second});
        adj[X.root(i.first.second)].push_back(
            {X.root(i.first.first), i.second});
      }
    }
    for (pair<pair<int, int>, int> i : A[w]) {
      if (X.same(i.first.first, i.first.second))
        ans[i.second] = "none";
      else {
        if (pirmas[X.root(i.first.first)] == 0) {
          dfs(X.root(i.first.first), -1015);
        }
      }
    }
    for (pair<pair<int, int>, int> i : A[w])
      X.merge(i.first.first, i.first.second);
    for (pair<pair<int, int>, int> i : A[w]) {
      adj[X.root(i.first.first)].clear();
      adj[X.root(i.first.second)].clear();
      for (int c : {i.first.first, i.first.second}) {
        pirmas[X.root(c)] = 0;
        maxi[X.root(c)] = 0;
      }
    }
  }
  for (int t = 0; t < tim; t++) cout << ans[t] << "\n";
}
