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;
struct Edge {
  int u, v, id, next;
  Edge() {}
  Edge(int a, int b, int c, int d) { u = a, v = b, id = c, next = d; }
};
vector<pair<pair<int, int>, pair<int, int> > > V;
Edge G[3 * 100000 + 10];
bool chk[100000 + 10];
int res[100000 + 10], st[100000 + 10], ed[100000 + 10], par[100000 + 10],
    component[100000 + 10], used[100000 + 10], cnt, EC = 0, head[100000 + 10];
void addedge(int u, int v, int id) {
  G[EC] = Edge(u, v, id, head[u]);
  head[u] = EC++;
}
int find_par(int x) {
  if (par[x] == x) return x;
  return par[x] = find_par(par[x]);
}
void dfs(int x, int prev) {
  st[x] = ed[x] = cnt++;
  chk[x] = true;
  for (int i = head[x]; i != -1; i = G[i].next) {
    int adj = G[i].v;
    if (G[i].id == prev) continue;
    if (!chk[adj]) {
      dfs(adj, G[i].id);
      ed[x] = min(ed[x], ed[adj]);
      if (st[x] < ed[adj]) {
        res[G[i].id] = 2;
      }
    } else
      ed[x] = min(ed[x], st[adj]);
  }
  used[x] = true;
}
int main() {
  int node, edge;
  cin >> node >> edge;
  for (int i = 0; i < edge; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    V.push_back(make_pair(make_pair(w, i), make_pair(u, v)));
  }
  sort(V.begin(), V.end());
  for (int i = 1; i <= node; i++) par[i] = i;
  memset(head, -1, sizeof head);
  for (int i = 0; i < edge; i++) {
    int j = i;
    while (j < edge && V[i].first.first == V[j].first.first) j++;
    for (int k = i; k < j; k++) {
      int u = find_par(V[k].second.first);
      int v = find_par(V[k].second.second);
      if (u != v) {
        addedge(u, v, V[k].first.second);
        addedge(v, u, V[k].first.second);
        res[V[k].first.second] = 1;
        component[k] = u;
      }
    }
    for (int k = i; k < j; k++) {
      if (!chk[component[k]] && component[k]) {
        cnt = 1;
        dfs(component[k], -1);
      }
    }
    for (int k = i; k < j; k++) {
      int u = find_par(V[k].second.first);
      int v = find_par(V[k].second.second);
      if (u != v) {
        par[u] = v;
        head[u] = head[v] = -1;
        component[k] = 0;
        chk[u] = chk[v] = false;
        used[u] = used[v] = false;
      }
    }
    i = j - 1;
  }
  for (int i = 0; i < edge; i++) {
    if (res[i] == 0)
      cout << "none" << endl;
    else if (res[i] == 1)
      cout << "at least one" << endl;
    else
      cout << "any" << endl;
  }
  return 0;
}
