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 maxN = 100022;
struct edge {
  edge(int a, int b, int i) {
    this->a = a;
    this->b = b;
    this->i = i;
  }
  int a, b, i;
};
vector<edge> E[10 * maxN];
int Baba[maxN], R[maxN];
pair<int, vector<edge>> C[maxN];
int getBaba(int v) { return Baba[v] < 0 ? v : Baba[v] = getBaba(Baba[v]); }
void marge(int u, int v) {
  if ((u = getBaba(u)) == (v = getBaba(v))) return;
  Baba[u] += Baba[v];
  Baba[v] = u;
}
int DFS(int v, int h, int pare = -1) {
  C[v].first = h;
  int r = h;
  for (auto e : C[v].second) {
    int u = getBaba(e.a) ^ getBaba(e.b) ^ v;
    if (e.i == pare) continue;
    if (!C[u].first) {
      int i = DFS(u, h + 1, e.i);
      R[e.i] = (i <= C[v].first ? 2 : 1);
      r = min(r, i);
    } else {
      r = min(r, C[u].first);
      R[e.i] = 2;
    }
  }
  return r;
}
void handle(int w) {
  vector<int> V;
  for (auto e : E[w]) {
    if (getBaba(e.a) == getBaba(e.b))
      R[e.i] = -1;
    else {
      V.push_back(getBaba(e.a));
      V.push_back(getBaba(e.b));
      C[getBaba(e.a)].second.push_back(e);
      C[getBaba(e.b)].second.push_back(e);
    }
  }
  for (auto v : V)
    if (C[v].first == 0) DFS(v, 1);
  for (auto e : E[w]) marge(e.a, e.b);
  for (auto v : V) {
    C[v].first = 0;
    C[v].second.clear();
  }
}
int main() {
  ios::sync_with_stdio(false);
  fill(Baba, Baba + maxN, -1);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int a, b, w;
    cin >> a >> b >> w;
    E[w].push_back(edge(a, b, i));
  }
  for (int i = 1; i < 10 * maxN; ++i) handle(i);
  for (int i = 0; i < m; ++i)
    if (R[i] == -1)
      cout << "none\n";
    else if (R[i] == 2)
      cout << "at least one\n";
    else
      cout << "any\n";
}
