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, fa[100005], ans[100005];
bool vis[100005];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
struct edge {
  int u, v, w, i;
  bool operator<(const edge x) const { return w < x.w; }
} e[100005];
vector<pair<int, int> > g[100005];
int low[100005], dfn[100005], idx;
void tarjan(int u) {
  low[u] = dfn[u] = ++idx;
  for (pair<int, int> i : g[u]) {
    int v = i.first;
    if (v != fa[v] || vis[i.second]) continue;
    vis[i.second] = 1;
    if (!dfn[v]) {
      tarjan(v);
      low[u] = min(low[u], low[v]);
      if (dfn[u] < low[v]) ans[i.second] = 1;
    } 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 = 1; i <= m; ++i)
    scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w), e[i].i = i;
  sort(e + 1, e + m + 1);
  for (int i = 1, j = 1, u, v; i <= m; i = j = j + 1) {
    while (j < m && e[j].w == e[j + 1].w) ++j;
    for (int k = i; k <= j; ++k) {
      u = find(e[k].u), v = find(e[k].v);
      if (u == v) {
        ans[e[k].i] = -1;
        continue;
      }
      g[u].push_back(make_pair(v, e[k].i));
      g[v].push_back(make_pair(u, e[k].i));
      dfn[u] = dfn[v] = 0;
    }
    for (int k = i; k <= j; ++k) {
      if (ans[e[k].i] == -1) continue;
      u = find(e[k].u), v = find(e[k].v);
      if (!dfn[u]) tarjan(u);
    }
    for (int k = i; k <= j; ++k) {
      if (ans[e[k].i] == -1) continue;
      u = find(e[k].u), v = find(e[k].v);
      if (u != v) fa[u] = v;
    }
  }
  for (int i = 1; i <= m; ++i)
    puts(!ans[i] ? "at least one" : (ans[i] > 0 ? "any" : "none"));
  return 0;
}
