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 inf = 1e9, maxn = 1e5 + 23, N = 1e6 + 20, LOG = 21, mod = 1e9 + 7;
const int dx[] = {0, -1, 1, 0}, dy[] = {1, 0, 0, -1};
const long double eps = 1e-13, pi = 3.14159265359;
const pair<unsigned long long, unsigned long long> base = make_pair(1171, 3307);
const long long INF = 1e18;
int timer, timer1, tin[maxn], fup[maxn], t[maxn];
int n, m, v[maxn], u[maxn], w[maxn], p[maxn], cnt[maxn], ans[maxn];
vector<int> ma[N], g[maxn];
set<int> q;
int get(int v) { return v == p[v] ? v : p[v] = get(p[v]); }
void merge(int v, int u) {
  v = get(v), u = get(u);
  if (cnt[v] < cnt[u]) swap(v, u);
  p[u] = v;
  cnt[v] += cnt[u];
}
void build() {
  for (int i = 0; i < n; ++i) {
    p[i] = i;
    cnt[i] = 1;
  }
}
void dfs(int vv, int p = -1) {
  t[vv] = timer;
  tin[vv] = fup[vv] = ++timer1;
  for (int i = 0; i < g[vv].size(); ++i) {
    int j = g[vv][i];
    if (j == p) continue;
    int to = vv == v[j] ? u[j] : v[j];
    if (t[to] == timer)
      fup[vv] = min(fup[vv], tin[to]);
    else {
      dfs(to, j);
      ans[j] = fup[to] > tin[vv] ? 2 : ans[j];
      fup[vv] = min(fup[vv], fup[to]);
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> v[i] >> u[i] >> w[i];
    --v[i], --u[i];
    ma[w[i]].push_back(i);
    q.insert(w[i]);
  }
  build();
  for (set<int>::iterator it = q.begin(); it != q.end(); ++it) {
    int cur = *it;
    ++timer;
    for (int i = 0; i < ma[cur].size(); ++i) {
      int j = ma[cur][i];
      v[j] = get(v[j]);
      u[j] = get(u[j]);
      g[v[j]].clear();
      g[u[j]].clear();
    }
    for (int i = 0; i < ma[cur].size(); ++i) {
      int j = ma[cur][i];
      if (v[j] != u[j]) {
        ans[j] = 1;
        g[v[j]].push_back(j);
        g[u[j]].push_back(j);
      }
    }
    for (int i = 0; i < ma[cur].size(); ++i) {
      int j = ma[cur][i];
      if (v[j] != u[j]) {
        if (t[v[j]] != timer) dfs(v[j]);
        if (t[u[j]] != timer) dfs(u[j]);
      }
    }
    for (int i = 0; i < ma[cur].size(); ++i) {
      int j = ma[cur][i];
      if (v[j] != u[j]) merge(v[j], u[j]);
    }
  }
  for (int i = 0; i < m; ++i)
    if (!ans[i])
      cout << "none\n";
    else if (ans[i] - 2)
      cout << "at least one\n";
    else
      cout << "any\n";
  return 0;
}
