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 = 100 * 1000 + 10;
struct edge {
  int x, y, w, id;
} e[maxn];
vector<pair<int, int> > g[maxn];
int mark[maxn], par[maxn], sz[maxn], ans[maxn], h[maxn];
bool cmp(edge a, edge b) {
  if (a.w == b.w) return a.id < b.id;
  return a.w < b.w;
}
int father(int v) {
  if (par[v] == 0) return v;
  return par[v] = father(par[v]);
}
void merge(int v, int u) {
  v = father(v), u = father(u);
  if (u == v) return;
  if (sz[v] < sz[u]) swap(v, u);
  par[u] = v;
  sz[v] += sz[u];
}
void dfs(int v, int par) {
  mark[v] = 1;
  int u, id, tmp = h[v];
  for (int i = 0; i < g[v].size(); i++) {
    u = g[v][i].first, id = g[v][i].second;
    if (id == par) continue;
    if (!mark[u]) {
      h[u] = h[v] + 1;
      dfs(u, id);
      if (h[u] > h[v]) ans[id] = 1;
    }
    tmp = min(tmp, h[u]);
  }
  h[v] = tmp;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++)
    cin >> e[i].x >> e[i].y >> e[i].w, e[i].id = i, ans[i] = 3;
  sort(e, e + m, cmp);
  e[m].w = -1;
  for (int i = 0; i < m;) {
    vector<edge> tmp;
    vector<int> nodes;
    tmp.push_back(e[i]), i++;
    while (e[i - 1].w == e[i].w) tmp.push_back(e[i]), i++;
    int v, u;
    for (int j = 0; j < tmp.size(); j++) {
      v = father(tmp[j].x), u = father(tmp[j].y);
      if (v == u) {
        ans[tmp[j].id] = 2;
        continue;
      }
      g[v].push_back(make_pair(u, tmp[j].id)),
          g[u].push_back(make_pair(v, tmp[j].id));
      nodes.push_back(v), nodes.push_back(u);
    }
    for (int j = 0; j < nodes.size(); j++) {
      int v = nodes[j];
      if (!mark[v]) h[v] = 0, dfs(v, -1);
    }
    for (int j = 0; j < nodes.size(); j++) {
      int v = nodes[j];
      g[v].clear(), mark[v] = 0;
    }
    for (int j = 0; j < tmp.size(); j++) merge(tmp[j].x, tmp[j].y);
  }
  for (int i = 0; i < m; i++)
    if (ans[i] == 2)
      cout << "none" << endl;
    else if (ans[i] == 1)
      cout << "any" << endl;
    else
      cout << "at least one" << endl;
  return 0;
}
