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;
void fast_io() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
typedef struct edge {
  int u, v, id;
} edge;
const int NONE = 0;
const int AT_LEAST_ONE = 1;
const int ANY = 2;
void make_set(vector<int> &parent, vector<int> &rank) {
  for (int i = 0; i < parent.size(); i++) {
    parent[i] = i;
    rank[i] = 0;
  }
}
int find(int a, vector<int> &parent) {
  if (parent[a] != a) parent[a] = find(parent[a], parent);
  return parent[a];
}
bool same(int a, int b, vector<int> &parent) {
  return find(a, parent) == find(b, parent);
}
void unite(int a, int b, vector<int> &parent, vector<int> &rank) {
  a = find(a, parent);
  b = find(b, parent);
  if (rank[a] <= rank[b]) {
    parent[a] = b;
    if (rank[a] == rank[b]) rank[b]++;
  } else
    parent[b] = a;
}
void tarjan(int u, int &T, map<int, vector<int>> &adjlist,
            map<int, vector<int>> &identifier, set<int> &processed,
            map<int, int> &disc, map<int, int> &low, vector<int> &ans) {
  if (low.count(u)) return;
  low[u] = disc[u] = ++T;
  for (int i = 0; i < adjlist[u].size(); i++) {
    int v = adjlist[u][i];
    int id = identifier[u][i];
    if (!processed.count(id)) {
      processed.insert(id);
      tarjan(v, T, adjlist, identifier, processed, disc, low, ans);
      low[u] = min(low[u], low[v]);
      if (low[v] > disc[u]) ans[id] = ANY;
    }
  }
}
int main() {
  fast_io();
  int n, m;
  int u, v, w;
  cin >> n >> m;
  map<int, vector<edge>> edgelist;
  vector<int> parent(n + 1), rank(n + 1), ans(m + 1, 0);
  for (int i = 0; i < m; i++) {
    cin >> u >> v >> w;
    if (!edgelist.count(w)) edgelist[w] = vector<edge>();
    edge e;
    e.u = u, e.v = v;
    e.id = (i + 1);
    edgelist[w].push_back(e);
  }
  make_set(parent, rank);
  for (auto it = edgelist.begin(); it != edgelist.end(); it++) {
    map<int, int> disc, low, ancestor;
    map<int, vector<int>> adjlist, identifier;
    set<int> processed;
    vector<edge> edges = it->second;
    for (int i = 0; i < edges.size(); i++) {
      int x = find(edges[i].u, parent);
      int y = find(edges[i].v, parent);
      if (!same(x, y, parent)) {
        ans[edges[i].id] = AT_LEAST_ONE;
        if (!adjlist.count(x)) adjlist[x] = vector<int>();
        adjlist[x].push_back(y);
        if (!adjlist.count(y)) adjlist[y] = vector<int>();
        adjlist[y].push_back(x);
        if (!identifier.count(x)) identifier[x] = vector<int>();
        identifier[x].push_back(edges[i].id);
        if (!identifier.count(y)) identifier[y] = vector<int>();
        identifier[y].push_back(edges[i].id);
      }
    }
    int T = 0;
    for (auto it_adj = adjlist.begin(); it_adj != adjlist.end(); it_adj++) {
      int vertex = it_adj->first;
      tarjan(vertex, T, adjlist, identifier, processed, disc, low, ans);
    }
    for (int i = 0; i < edges.size(); i++) {
      int x = find(edges[i].u, parent);
      int y = find(edges[i].v, parent);
      if (!same(x, y, parent)) unite(x, y, parent, rank);
    }
  }
  for (int i = 1; i <= m; i++) {
    string res;
    if (ans[i] == 0)
      res = "none";
    else if (ans[i] == 1)
      res = "at least one";
    else
      res = "any";
    cout << res << '\n';
  }
  return 0;
}
