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 = 1e5, MAX_M = 1e5, MAX_W = 1e6;
vector<pair<int, int> > G[MAX_N];
int N, M;
struct edge {
  int a, b, id, w;
};
bool operator<(edge a, edge b) { return a.w < b.w; }
edge E[MAX_M];
int ufpar[MAX_N];
int ans[MAX_M];
int fndpar(int v) {
  if (ufpar[v] == v) {
    return v;
  }
  return ufpar[v] = fndpar(ufpar[v]);
}
void unite(int u, int v) {
  u = fndpar(u);
  v = fndpar(v);
  ufpar[u] = v;
}
int vis[MAX_N], par[MAX_N], lest[MAX_N], pare[MAX_N];
int t;
void dfs(int v, int pe) {
  vis[v] = t;
  lest[v] = t;
  t++;
  for (auto e : G[v]) {
    if (vis[e.first] == -1) {
      par[e.first] = v;
      pare[e.first] = e.second;
      dfs(e.first, e.second);
      lest[v] = min(lest[v], lest[e.first]);
    } else if (e.second != pe) {
      lest[v] = min(lest[v], vis[e.first]);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  for (int i = (0); i < (M); i++) {
    int a, b, w;
    cin >> a >> b >> w;
    a--;
    b--;
    E[i] = (edge){a, b, i, w};
  }
  for (int i = (0); i < (N); i++) ufpar[i] = i;
  sort(E, E + M);
  for (int i = (0); i < (M); i++) {
    int cst = E[i].w;
    int i2 = i;
    while (i2 < M && E[i2].w == cst) i2++;
    set<int> vs;
    for (int j = i; j < i2; j++) {
      edge e = E[j];
      int u = fndpar(e.a), v = fndpar(e.b);
      if (u == v) {
        ans[e.id] = 2;
      } else {
        ans[e.id] = 1;
        if (vs.find(u) == vs.end()) {
          G[u].clear();
        }
        if (vs.find(v) == vs.end()) {
          G[v].clear();
        }
        vs.insert(u);
        vs.insert(v);
        G[u].push_back(pair<int, int>(v, e.id));
        G[v].push_back(pair<int, int>(u, e.id));
      }
    }
    for (auto v : vs) {
      vis[v] = -1;
      par[v] = -1;
      pare[v] = -1;
    }
    t = 0;
    for (auto v : vs) {
      if (vis[v] == -1) {
        dfs(v, -1);
      }
    }
    for (auto v : vs) {
      int p = par[v];
      if (p != -1 && lest[v] > vis[p]) {
        ans[pare[v]] = 0;
      }
    }
    for (int j = i; j < i2; j++) {
      edge e = E[j];
      unite(e.a, e.b);
    }
    i = i2 - 1;
  }
  for (int i = (0); i < (M); i++) {
    switch (ans[i]) {
      case 0:
        cout << "any";
        break;
      case 1:
        cout << "at least one";
        break;
      case 2:
        cout << "none";
        break;
    }
    cout << endl;
  }
}
