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;
int n, m, in1, in2, in3, siz, opt, ccn;
int op[100001], fa[100001];
int hav[100001];
vector<pair<int, pair<int, int> > > ed[100001][2];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  scanf("%d%d", &n, &m);
  if (n == m + 1) {
    for (int i = 1; i <= m; i++) {
      printf("any\n");
    }
    return 0;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &in1, &in2, &in3);
    ed[in1][opt].push_back(
        pair<int, pair<int, int> >(in3, pair<int, int>(in2, i)));
    ed[in2][opt].push_back(
        pair<int, pair<int, int> >(in3, pair<int, int>(in1, i)));
  }
  for (int i = 1; i <= n; i++) {
    fa[i] = i;
  }
  while (true) {
    if (siz == 1) break;
    ccn++;
    for (int i = 1; i <= n; i++) {
      if (ed[i][opt].empty()) continue;
      if (ed[i][opt].size() == 1) {
        op[ed[i][opt][0].second.second] = 1;
        if (find(i) != find(ed[i][opt][0].second.first))
          fa[fa[i]] = fa[ed[i][opt][0].second.first];
        continue;
      }
      sort(ed[i][opt].begin(), ed[i][opt].end());
      if (ed[i][opt][0].first != ed[i][opt][1].first) {
        op[ed[i][opt][0].second.second] = 1;
        if (find(i) != find(ed[i][opt][0].second.first))
          fa[fa[i]] = fa[ed[i][opt][0].second.first];
        continue;
      }
    }
    if (ccn >= 300) {
      for (int i = 1; i <= n; i++) {
        if (ed[i][opt].empty()) continue;
        if (ed[i][opt].size() == 1) continue;
        vector<pair<int, pair<int, int> > >::iterator ite = ed[i][opt].begin(),
                                                      ite1 = ite + 1;
        if (op[ite->second.second] == 1) continue;
        while (ite1 != ed[i][opt].end() && ite->first == ite1->first) {
          if (op[ite1->second.second] == 1) goto hh;
          ite++;
          ite1++;
        }
        ite = ed[i][opt].begin(), ite1 = ite + 1;
        op[ite->second.second] = 2;
        if (find(i) != find(ite->second.first))
          fa[fa[i]] = fa[ite->second.first];
        while (ite1 != ed[i][opt].end() && ite->first == ite1->first) {
          op[ite1->second.second] = 2;
          if (find(i) != find(ite1->second.first))
            fa[fa[i]] = fa[ite1->second.first];
          ite++;
          ite1++;
        }
      hh:;
      }
      break;
    }
    memset(hav, 0, sizeof(hav));
    siz = 0;
    for (int i = 1; i <= n; i++) {
      if (!hav[find(i)]) {
        hav[find(i)] = ++siz;
      }
    }
    for (int i = 1; i <= siz; i++) {
      ed[i][opt ^ 1].clear();
    }
    for (int i = 1; i <= n; i++) {
      for (vector<pair<int, pair<int, int> > >::iterator ite =
               ed[i][opt].begin();
           ite != ed[i][opt].end(); ite++) {
        if (find(i) != find(ite->second.first)) {
          ed[hav[find(i)]][opt ^ 1].push_back(pair<int, pair<int, int> >(
              ite->first, pair<int, int>(hav[find(ite->second.first)],
                                         ite->second.second)));
        }
      }
    }
    for (int i = 1; i <= siz; i++) {
      fa[i] = i;
    }
    n = siz;
    opt ^= 1;
  }
  for (int i = 1; i <= m; i++) {
    if (op[i] == 1) printf("any\n");
    if (op[i] == 2) printf("at least one\n");
    if (op[i] == 0) printf("none\n");
  }
  return 0;
}
