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, x[100001], y[100001], z[100001], p[100001];
int ans[100001], f[100001], h[100001], pe[100001], d[100001];
vector<pair<int, int> > v[100001];
bool cmp(const int &x, const int &y) { return z[x] < z[y]; }
int par(int x) {
  while (pe[x]) x = pe[x];
  return x;
}
void uni(int x, int y) {
  x = par(x);
  y = par(y);
  v[x].clear();
  v[y].clear();
  f[x] = 0;
  f[y] = 0;
  if (x == y) return;
  if (h[x] > h[y])
    pe[y] = x;
  else {
    pe[x] = y;
    if (h[x] == h[y]) h[y]++;
  }
}
void add_edge(int x, int y, int i) {
  if (x == y) {
    return;
  }
  ans[i] = 1;
  v[x].push_back({y, i});
  v[y].push_back({x, i});
}
void kruskal(int c, int g, int h) {
  f[c] = true;
  d[c] = h;
  for (pair<int, int> i : v[c])
    if (!f[i.first]) {
      kruskal(i.first, i.second, h + 1);
      d[c] = min(d[c], d[i.first]);
    } else if (i.second != g)
      d[c] = min(d[c], d[i.first]);
  if (d[c] == h) ans[g] = 2;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x[i] >> y[i] >> z[i];
    p[i] = i;
  }
  sort(p + 1, p + m + 1, cmp);
  for (int i = 1; i <= m;) {
    int j;
    for (int j = i; z[p[j]] == z[p[i]]; j++)
      add_edge(par(x[p[j]]), par(y[p[j]]), p[j]);
    for (j = i; z[p[j]] == z[p[i]]; j++) {
      int k = par(x[p[j]]);
      if (!f[k]) kruskal(k, 0, 0);
    }
    for (j = i; z[p[j]] == z[p[i]]; j++) uni(x[p[j]], y[p[j]]);
    i = j;
  }
  for (int i = 1; i <= m; i++) {
    if (!ans[i])
      cout << "none" << endl;
    else if (ans[i] == 1)
      cout << "at least one" << endl;
    else
      cout << "any" << endl;
  }
  return 0;
}
