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;
const long long mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double PI = acos(-1);
int n;
int ans[maxn];
int p[maxn];
int fa[maxn];
struct E {
  int from;
  int to;
  int nxt;
  int l;
  int id;
  bool cut;
  bool operator<(const E &e) const { return l < e.l; }
};
vector<E> ee;
E edge[2 * maxn];
int dfn[maxn];
int low[maxn];
int ecnt = 0;
int vis[maxn];
void add_edge(int u, int v, int id) {
  E e;
  e.to = v;
  e.nxt = p[u];
  e.id = id;
  e.cut = false;
  edge[ecnt] = e;
  p[u] = ecnt;
  ecnt++;
}
int getfa(int u) {
  if (fa[u] == u) return u;
  return fa[u] = getfa(fa[u]);
}
int root;
void dfs(int u, int fa, int dep) {
  vis[u] = 1;
  dfn[u] = low[u] = dep;
  int tot = 0;
  for (int i = p[u]; i != -1; i = edge[i].nxt) {
    E e = edge[i];
    if (e.id == fa) continue;
    if (vis[e.to] == 1) {
      low[u] = min(low[u], dfn[e.to]);
    } else if (0 == vis[e.to]) {
      tot++;
      dfs(e.to, e.id, dep + 1);
      if (low[u] > low[e.to]) {
        low[u] = low[e.to];
      }
      if (low[e.to] > dfn[u]) {
        edge[i].cut = true;
        edge[i ^ 1].cut = true;
      }
    }
  }
  vis[u] = 2;
}
void solve() {
  sort(ee.begin(), ee.end());
  vector<int> tmp;
  for (int i = 0; i < ee.size(); i++) {
    tmp.push_back(i);
    if (i < ee.size() - 1 && ee[i].l == ee[i + 1].l) {
      continue;
    } else {
      for (int j = 0; j < tmp.size(); j++) {
        E e = ee[tmp[j]];
        int u = e.from;
        int v = e.to;
        u = getfa(u);
        v = getfa(v);
        if (u == v) {
          ans[e.id] = 0;
        } else {
          add_edge(u, v, e.id);
          add_edge(v, u, e.id);
          vis[u] = false;
          vis[v] = false;
        }
      }
      for (int j = 0; j < tmp.size(); j++) {
        int u = ee[tmp[j]].from;
        u = getfa(u);
        if (!vis[u]) {
          root = u;
          dfs(u, -1, 0);
        }
      }
      for (int j = 0; j < ecnt; j++) {
        if (edge[j].cut) {
          ans[edge[j].id] = 2;
        } else
          ans[edge[j].id] = 1;
      }
      for (int j = 0; j < tmp.size(); j++) {
        int u = ee[tmp[j]].from;
        int v = ee[tmp[j]].to;
        u = getfa(u);
        v = getfa(v);
        p[u] = -1;
        p[v] = -1;
      }
      for (int j = 0; j < tmp.size(); j++) {
        int u = ee[tmp[j]].from;
        int v = ee[tmp[j]].to;
        u = getfa(u);
        v = getfa(v);
        fa[v] = u;
      }
      ecnt = 0;
      tmp.clear();
    }
  }
}
int m;
int main() {
  while (cin >> n >> m) {
    memset(p, -1, sizeof(p));
    ecnt = 0;
    for (int i = 0; i < maxn; i++) fa[i] = i;
    for (int i = 0; i < m; i++) {
      int u, v, l;
      scanf("%d%d%d", &u, &v, &l);
      ee.push_back((E){u, v, 0, l, i});
    }
    solve();
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 0)
      cout << "none" << endl;
    else if (ans[i] == 1)
      cout << "at least one" << endl;
    else
      cout << "any" << endl;
  }
}
