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, w, ind;
};
const int N = 1000 * 100 + 5;
EDGE edge[N];
vector<pair<int, int> > adj[N];
short ans[N];
int st[N], num[N], par[N], siz[N], k, stt;
bool mark[N], mark3[N];
void IsCutedge(int v, int parr, int ind) {
  mark3[v] = 1;
  st[v] = stt++;
  bool flag = 0;
  int kk = 0;
  for (int i = 0; i < (int)adj[v].size(); i++) {
    int u = adj[v][i].first, indexx = adj[v][i].second;
    if (!mark3[u]) {
      kk += k;
      k = 0;
      IsCutedge(u, v, indexx);
    } else {
      if ((u == parr && !flag) || st[u] > st[v])
        int temp = 0;
      else {
        num[v]++;
        num[u]--;
      }
      if (u == parr) flag = 1;
    }
  }
  k += num[v] + kk;
  if (k == 0 && ans[ind] != 1 && v != parr) ans[ind] = 3;
  return;
}
int findpar(int v) {
  if (par[v] == -1) return v;
  return par[v] = findpar(par[v]);
}
void merge(int a, int b) {
  a = findpar(a);
  b = findpar(b);
  if (a == b) return;
  if (siz[a] > siz[b]) swap(a, b);
  par[a] = b;
  siz[b] += siz[a];
  return;
}
void clearr(int t) {
  adj[t].clear();
  mark3[t] = 0;
  st[t] = 0;
  num[t] = 0;
  return;
}
bool comp(EDGE a, EDGE b) {
  if (a.w < b.w) return 1;
  return 0;
}
int main() {
  int n, m, x, y, w;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> edge[i].v >> edge[i].u >> edge[i].w;
    edge[i].ind = i;
  }
  sort(edge, edge + m, comp);
  fill(par, par + n, -1);
  fill(siz, siz + n, 1);
  for (int i = 0; i <= m; i++) {
    EDGE e = edge[i];
    int u = e.u, v = e.v, ind = e.ind;
    if ((i && e.w != edge[i - 1].w) || i == m) {
      for (int j = i - 1; j >= 0 && !mark[j]; j--) {
        int x = findpar(edge[j].u), y = findpar(edge[j].v);
        adj[x].push_back(make_pair(y, edge[j].ind));
        adj[y].push_back(make_pair(x, edge[j].ind));
      }
      for (int j = i - 1; j >= 0 && !mark[j]; j--) {
        int p = findpar(edge[j].u);
        if (!mark3[p]) {
          IsCutedge(p, p, 0);
          k = 0;
        }
      }
      for (int j = i - 1; j >= 0 && mark[j] == 0; j--) {
        int uu = findpar(edge[j].u);
        int vv = findpar(edge[j].v);
        clearr(uu);
        clearr(vv);
      }
      for (int j = i - 1; j >= 0 && !mark[j]; j--) {
        mark[j] = 1;
        merge(edge[j].v, edge[j].u);
      }
    }
    if (i < m) {
      if (findpar(u) == findpar(v))
        ans[ind] = 1;
      else
        ans[ind] = 2;
    }
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 1)
      cout << "none\n";
    else if (ans[i] == 2)
      cout << "at least one\n";
    else
      cout << "any\n";
  }
  return 0;
}
