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 int N = 1e5 + 5;
struct Edge {
  int id, u, v, w, st;
  Edge() {}
  Edge(int _id, int _u, int _v, int _w, int _st) {
    id = _id;
    u = _u;
    v = _v;
    w = _w;
    st = _st;
  }
  bool operator<(const Edge &a) const { return w < a.w; }
} a[N];
vector<pair<int, int> > G[N];
int n, m, fa[N], dfn[N], low[N], dfs_clock;
bool cmp(const Edge &a, const Edge &b) { return a.id < b.id; }
int find(int x) {
  if (fa[x] != x) fa[x] = find(fa[x]);
  return fa[x];
}
void dfs(int u, int fa_id) {
  low[u] = dfn[u] = ++dfs_clock;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i].first, id = G[u][i].second;
    if (id == fa_id) continue;
    if (!dfn[v]) {
      dfs(v, id);
      low[u] = min(low[u], low[v]);
      if (low[v] > dfn[u]) a[id].st = 2;
    } else
      low[u] = min(low[u], dfn[v]);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    a[i] = Edge(i + 1, u, v, w, 0);
  }
  sort(a, a + m);
  for (int i = 0; i < m; i++) {
    int s = i, e = i;
    while (e + 1 < m && a[e + 1].w == a[s].w) e++;
    for (int j = s; j <= e; j++) {
      int x = find(a[j].u), y = find(a[j].v);
      if (x == y) continue;
      G[x].push_back(make_pair(y, j));
      G[y].push_back(make_pair(x, j));
      a[j].st = 1;
    }
    for (int j = s; j <= e; j++) {
      int x = find(a[j].u), y = find(a[j].v);
      if (x == y || dfn[x]) continue;
      dfs(x, -1);
    }
    for (int j = s; j <= e; j++) {
      int x = find(a[j].u), y = find(a[j].v);
      if (x == y) continue;
      dfn[x] = dfn[y] = low[x] = low[y] = 0;
      G[x].clear();
      G[y].clear();
      fa[x] = y;
    }
    i = e;
  }
  sort(a, a + m, cmp);
  for (int i = 0; i < m; i++)
    if (a[i].st == 0)
      puts("none");
    else if (a[i].st == 1)
      puts("at least one");
    else
      puts("any");
  return 0;
}
