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 = 100000 + 10;
int n, m;
struct Edge {
  int a, b, w;
  int id;
  void read() {
    scanf("%d%d%d", &a, &b, &w);
    --a, --b;
  }
  bool operator<(const Edge& e) const { return w < e.w; }
};
Edge es[MAX_N];
int F[MAX_N];
int fd(int x) { return F[x] == x ? x : F[x] = fd(F[x]); }
string ans[MAX_N];
vector<pair<int, int> > E[MAX_N];
int ord[MAX_N], low[MAX_N];
int mark[MAX_N];
int cur, pt;
void dfs(int u, int pid) {
  ord[u] = low[u] = pt++;
  mark[u] = cur;
  for (auto e : E[u]) {
    int t = e.first, id = e.second;
    if (id == pid) continue;
    if (mark[t] != cur) {
      dfs(t, id);
      low[u] = min(low[u], low[t]);
    } else {
      low[u] = min(low[u], ord[t]);
    }
  }
  if (pid != -1) {
    if (low[u] == ord[u]) {
      ans[pid] = "any";
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    es[i].read();
    es[i].id = i;
  }
  for (int i = 0; i < n; ++i) {
    F[i] = i;
  }
  sort(es, es + m);
  for (int i = 0, j; i < m; i = j) {
    for (j = i; j < m && es[j].w == es[i].w; ++j)
      ;
    for (int k = i; k < j; ++k) {
      Edge& e = es[k];
      if (fd(e.a) == fd(e.b))
        ans[e.id] = "none";
      else
        ans[e.id] = "at least one";
      E[fd(e.a)].clear();
      E[fd(e.b)].clear();
    }
    for (int k = i; k < j; ++k) {
      Edge& e = es[k];
      int a = fd(e.a), b = fd(e.b);
      if (a != b) {
        E[a].push_back(make_pair(b, e.id));
        E[b].push_back(make_pair(a, e.id));
      }
    }
    ++cur;
    pt = 0;
    for (int k = i; k < j; ++k) {
      Edge& e = es[k];
      int a = fd(e.a), b = fd(e.b);
      if (a != b) {
        if (mark[a] != cur) {
          dfs(a, -1);
        }
        if (mark[b] != cur) {
          dfs(b, -1);
        }
      }
    }
    for (int k = i; k < j; ++k) {
      Edge& e = es[k];
      int a = fd(e.a), b = fd(e.b);
      F[a] = b;
    }
  }
  for (int i = 0; i < m; ++i) {
    puts(ans[i].c_str());
  }
  return 0;
}
