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 = 1e5 + 10;
vector<int> edge;
vector<pair<int, int>> G[maxn];
int a[maxn], b[maxn], c[maxn], ans[maxn], tin[maxn], low[maxn], t;
bool v[maxn];
struct dsu {
  int p[maxn];
  void init() {
    for (int i = 0; i < maxn; ++i) p[i] = i;
  }
  int find(int x) { return x == p[x] ? x : p[x] = find(p[x]); }
  void merge(int x, int y) { p[find(x)] = find(y); }
} dsu;
void dfs(int now) {
  tin[now] = low[now] = ++t;
  for (auto u : G[now])
    if (!v[u.second]) {
      v[u.second] = true;
      if (tin[u.first]) {
        low[now] = min(low[now], tin[u.first]);
        continue;
      }
      dfs(u.first);
      low[now] = min(low[now], low[u.first]);
      if (low[u.first] > tin[now]) ans[u.second] = 1;
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a[i] >> b[i] >> c[i];
    edge.push_back(i);
  }
  sort(edge.begin(), edge.end(),
       [&](const int &i, const int &j) { return c[i] < c[j]; });
  dsu.init();
  fill(ans, ans + maxn, -1);
  for (size_t i = 0; i < edge.size();) {
    size_t ptr = i;
    while (ptr < edge.size() && c[edge[ptr]] == c[edge[i]]) {
      int u = dsu.find(a[edge[ptr]]), v = dsu.find(b[edge[ptr]]);
      if (u == v)
        ans[edge[ptr]] = 0;
      else
        G[u].emplace_back(v, edge[ptr]), G[v].emplace_back(u, edge[ptr]);
      ++ptr;
    }
    for (size_t j = i; j < ptr; ++j) {
      if (!v[edge[j]]) dfs(dsu.find(a[edge[j]]));
    }
    for (size_t j = i; j < ptr; ++j) {
      int u = dsu.find(a[edge[j]]), v = dsu.find(b[edge[j]]);
      tin[u] = low[u] = tin[v] = low[v] = 0;
      G[u].clear();
      G[v].clear();
      dsu.merge(u, v);
    }
    i = ptr;
  }
  for (int i = 0; i < m; ++i) {
    if (ans[i] == 0)
      cout << "none" << endl;
    else if (ans[i] == 1)
      cout << "any" << endl;
    else
      cout << "at least one" << endl;
  }
  return 0;
}
