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;
class DSet {
 public:
  static const int SIZE = 1000005;
  int sz[SIZE], p[SIZE];
  void init(int n) {
    for (int i = 0; i < n; i += sz[p[i] = i] = 1)
      ;
  }
  int root(int x) { return p[x] != x ? p[x] = root(p[x]) : x; }
  bool join(int x, int y) {
    x = root(x), y = root(y);
    if (x == y) return false;
    if (sz[x] <= sz[y])
      sz[p[x] = y] += sz[x];
    else
      sz[p[y] = x] += sz[y];
    return true;
  }
} ds;
const char* msg[] = {"none", "any", "at least one"};
tuple<int, int, int, int> e[100000];
map<int, vector<pair<int, int>>> arc;
int tag[100000], low[100000], ans[100000], cnt;
void tarjan(int x, int use) {
  tag[x] = low[x] = ++cnt;
  for (auto c : arc[x]) {
    int y = c.first, o = c.second;
    if (o == use) continue;
    if (tag[y]) {
      low[x] = min(low[x], tag[y]);
    } else {
      tarjan(y, o);
      low[x] = min(low[x], low[y]);
      if (low[y] > tag[x]) ans[o] = 1;
    }
  }
}
int main() {
  int n, m, x, y, c, o;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &x, &y, &c);
    e[i] = tie(c, --x, --y, i);
  }
  sort(e, e + m);
  ds.init(n);
  vector<int> u;
  for (int i = 0; i < m; i++) {
    u.push_back(i);
    if (get<0>(e[i]) == get<0>(e[i + 1])) continue;
    for (auto r : u) {
      tie(c, x, y, o) = e[r];
      x = ds.root(x);
      y = ds.root(y);
      if (x == y) continue;
      ans[o] = 2;
      arc[x].push_back({y, o});
      arc[y].push_back({x, o});
    }
    for (auto t : arc) {
      if (tag[t.first]) continue;
      tarjan(t.first, -1);
      cnt = 0;
    }
    for (auto t : arc) tag[t.first] = low[t.first] = 0;
    arc.clear();
    for (auto r : u) ds.join(get<1>(e[r]), get<2>(e[r]));
    u.clear();
  }
  for (int i = 0; i < m; i++) puts(msg[ans[i]]);
}
