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;
vector<int> edges[1000005];
vector<int> graph[100005];
int dsu[100005];
int u[100005], v[100005];
int ans[100005];
int dfn[100005], low[100005], t;
inline int getrt(int n) { return dsu[n] = (dsu[n] == n ? n : getrt(dsu[n])); }
inline void dfs(int n, int p) {
  dfn[n] = low[n] = ++t;
  for (int e : graph[n]) {
    if (e == p) {
      continue;
    }
    if (v[e] == n) {
      swap(u[e], v[e]);
    }
    if (!dfn[v[e]]) {
      dfs(v[e], e);
    }
    low[n] = min(low[n], low[v[e]]);
  }
  if (dfn[n] == low[n]) {
    ans[p] = 3;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  for (int i = 1; i <= M; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    edges[c].emplace_back(i);
    u[i] = a, v[i] = b;
  }
  for (int i = 1; i <= N; i++) {
    dsu[i] = i;
  }
  for (int k = 1; k <= 1000000; k++) {
    for (int e : edges[k]) {
      u[e] = getrt(u[e]), v[e] = getrt(v[e]);
      if (u[e] == v[e]) {
        ans[e] = 1;
      } else {
        int x = u[e], y = v[e];
        graph[x].emplace_back(e);
        graph[y].emplace_back(e);
        dfn[x] = dfn[y] = 0;
        low[x] = low[y] = 0;
      }
    }
    t = 0;
    for (int e : edges[k]) {
      if (!dfn[u[e]]) {
        dfs(u[e], 0);
      }
    }
    for (int e : edges[k]) {
      if (!ans[e]) {
        ans[e] = 2;
      }
      if (getrt(u[e]) != getrt(v[e])) {
        dsu[getrt(u[e])] = getrt(v[e]);
      }
      graph[u[e]].clear();
      graph[v[e]].clear();
    }
  }
  for (int i = 1; i <= M; i++) {
    if (ans[i] == 1) {
      cout << "none\n";
    } else if (ans[i] == 2) {
      cout << "at least one\n";
    } else {
      cout << "any\n";
    }
  }
}
