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 max_n = 100110;
vector<int> adj[max_n];
vector<int> adj1[max_n];
int d[max_n];
int perm[max_n];
int h[max_n];
int p[max_n];
pair<int, int> e[max_n];
bool mark[max_n];
int ans[max_n];
const int inf = 100000000;
int n, m;
bool f(const int &A, const int &B) { return d[A] < d[B]; }
void init() {
  for (int i = 1; i <= n; i++) p[i] = i;
}
int find(int v) {
  if (v != p[v]) p[v] = find(p[v]);
  return p[v];
}
void uni(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return;
  if (h[u] < h[v])
    p[u] = v;
  else {
    if (h[u] == h[v]) h[u]++;
    p[v] = u;
  }
}
int dfs(int v, int pos1) {
  mark[v] = true;
  int min1 = inf;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    int pos = adj1[v][i];
    if (u == v) {
      ans[pos] = 3;
    } else if (mark[u] && pos != pos1) {
      ans[pos] = 2;
      min1 = min(min1, h[u]);
    } else if (!mark[u]) {
      h[u] = h[v] + 1;
      min1 = min(dfs(u, pos), min1);
    }
  }
  if (pos1 != -1) {
    if (min1 < h[v])
      ans[pos1] = 2;
    else
      ans[pos1] = 1;
  }
  return min1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  init();
  for (int i = 0; i < m; i++) {
    perm[i] = i;
    cin >> e[i].first >> e[i].second >> d[i];
  }
  sort(perm, perm + m, f);
  for (int i = 0; i < m; i++) {
    int tmp = i;
    while (d[perm[tmp]] == d[perm[i]] && tmp < m) {
      int pos = perm[tmp];
      int u = e[pos].first;
      int v = e[pos].second;
      int u1 = find(u);
      int v1 = find(v);
      adj[v1].push_back(u1);
      adj[u1].push_back(v1);
      adj1[u1].push_back(pos);
      adj1[v1].push_back(pos);
      tmp++;
    }
    tmp = i;
    while (d[perm[tmp]] == d[perm[i]] && tmp < m) {
      int pos = perm[tmp];
      int u = e[pos].first;
      u = find(u);
      if (!mark[u]) {
        dfs(u, -1);
      }
      tmp++;
    }
    tmp = i;
    while (d[perm[tmp]] == d[perm[i]] && tmp < m) {
      int pos = perm[tmp];
      int u = e[pos].first;
      int v = e[pos].second;
      u = find(u);
      v = find(v);
      adj[u].clear();
      adj[v].clear();
      adj1[u].clear();
      adj1[v].clear();
      mark[u] = mark[v] = false;
      h[v] = h[u] = 0;
      tmp++;
    }
    tmp = i;
    while (d[perm[tmp]] == d[perm[i]] && tmp < m) {
      int pos = perm[tmp];
      int v = e[pos].first;
      int u = e[pos].second;
      uni(u, v);
      tmp++;
    }
    i = tmp - 1;
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 1)
      cout << "any" << endl;
    else if (ans[i] == 2)
      cout << "at least one" << endl;
    else if (ans[i] == 3)
      cout << "none" << endl;
  }
  return 0;
}
