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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
struct my {
  int t1, t2;
  int c;
  int id;
  int i;
};
bool cmp(my a, my b) { return a.c < b.c; }
bool cmpi(my a, my b) { return a.i < b.i; }
int parent[100013];
int rrank[100013];
void init(int n) {
  for (int i = 0; i < n; i++) {
    parent[i] = i;
    rrank[i] = 1;
  }
}
int find_set(int a) {
  if (a == parent[a]) {
    return a;
  }
  parent[a] = find_set(parent[a]);
  return parent[a];
}
void union_set(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a == b) return;
  if (rrank[a] == rrank[b]) {
    parent[a] = b;
    rrank[a]++;
    return;
  }
  if (rrank[a] < rrank[b]) {
    parent[a] = b;
  } else {
    parent[b] = a;
  }
  return;
}
int main() {
  int n, m;
  cin >> n >> m;
  init(n);
  my a[100013];
  for (int i = 0; i < m; i++) {
    cin >> a[i].t1 >> a[i].t2 >> a[i].c;
    a[i].t1--;
    a[i].t2--;
    a[i].i = i;
    a[i].id = 0;
  }
  sort(a, a + m, cmp);
  for (int i = 0; i < m;) {
    int j = i;
    map<int, set<int> > t;
    while (j < m && a[i].c == a[j].c) {
      if (find_set(a[j].t1) != find_set(a[j].t2)) {
        a[j].id++;
        t[find_set(a[j].t1)].insert(j);
        t[find_set(a[j].t2)].insert(j);
      }
      j++;
    }
    for (int k = i; k < j; k++) {
      if (a[k].id == 1) {
        if (t[find_set(a[k].t1)].size() == 1 ||
            t[find_set(a[k].t2)].size() == 1) {
          int kk = k;
          while (true) {
            if (t[find_set(a[kk].t1)].size() == 1) {
              a[kk].id++;
              t[find_set(a[kk].t1)].erase(kk);
              t[find_set(a[kk].t2)].erase(kk);
              if (t[find_set(a[kk].t2)].size() == 1) {
                kk = *(t[find_set(a[kk].t2)].begin());
              } else {
                break;
              }
            } else if (t[find_set(a[kk].t2)].size() == 1) {
              a[kk].id++;
              t[find_set(a[kk].t2)].erase(kk);
              t[find_set(a[kk].t1)].erase(kk);
              if (t[find_set(a[kk].t1)].size() == 1) {
                kk = *(t[find_set(a[kk].t1)].begin());
              } else {
                break;
              }
            }
          }
        }
      }
    }
    for (int k = i; k < j; k++) {
      union_set(a[k].t1, a[k].t2);
    }
    i = j;
  }
  sort(a, a + m, cmpi);
  for (int i = 0; i < m; i++) {
    if (a[i].id == 0) {
      cout << "none\n";
    } else {
      if (a[i].id == 1) {
        cout << "at least one\n";
      } else {
        cout << "any\n";
      }
    }
  }
  return 0;
}
