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 = 1000 * 1000 + 10;
const int maxn5 = 1000 * 100 + 10;
const int maxn3 = 1000 + 10;
const long long mod = 1000 * 1000 * 1000 + 7;
const long long inf = 1LL * 2 * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 10;
int par[maxn5], a[maxn5], b[maxn5], rnk[maxn5], h[maxn5], mn[maxn5], pos[maxn5];
bool mark[maxn5];
vector<int> ver, av[maxn];
vector<pair<int, int>> adj[maxn5];
int get_par(int a) { return (par[a] == 0 ? a : par[a] = get_par(par[a])); }
void join(int a, int b) {
  a = get_par(a);
  b = get_par(b);
  if (a == b) return;
  if (rnk[a] > rnk[b]) swap(a, b);
  if (rnk[a] == rnk[b]) rnk[b]++;
  par[a] = b;
  return;
}
void dfs(int v, int par = -1) {
  mark[v] = true;
  mn[v] = v;
  for (auto [u, ind] : adj[v]) {
    if (!mark[u]) {
      h[u] = h[v] + 1;
      dfs(u, ind);
      if (h[mn[u]] > h[v]) {
        pos[ind] = 2;
      } else if (h[mn[u]] < h[mn[v]])
        mn[v] = mn[u];
    } else if (ind != par and h[mn[u]] < h[mn[v]])
      mn[v] = mn[u];
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int len;
    cin >> a[i] >> b[i] >> len;
    av[len].push_back(i);
  }
  for (int i = 1; i < maxn; i++)
    if (!av[i].empty()) {
      for (auto ind : av[i]) {
        int x = get_par(a[ind]), y = get_par(b[ind]);
        if (x != y) {
          adj[x].push_back({y, ind}), adj[y].push_back({x, ind});
          ver.push_back(x);
          ver.push_back(y);
          pos[ind] = 1;
        }
      }
      for (auto u : ver)
        if (!mark[u]) {
          h[u] = 0;
          dfs(u);
        }
      for (auto u : ver) {
        mark[u] = false;
        adj[u].clear();
      }
      ver.clear();
      for (auto ind : av[i]) join(a[ind], b[ind]);
    }
  for (int i = 0; i < m; i++)
    cout << (pos[i] == 0 ? "none" : (pos[i] == 1 ? "at least one" : "any"))
         << '\n';
  return 0;
}
