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 = 100000 + 3;
const char* inf[] = {"none", "any", "at least one"};
int f[N], dfn[N], low[N];
int n, m, idx;
int res[N];
vector<pair<int, int> > p[N];
struct Tuple {
  int x, y, d, idx;
  bool operator<(const Tuple& t) const { return d < t.d; }
} msg[N];
void init() {
  for (int i = 1; i <= n; ++i) f[i] = i;
}
int root(int x) { return x == f[x] ? x : f[x] = root(f[x]); }
void unions(int x, int y) {
  x = root(x);
  y = root(y);
  if (x == y) return;
  if (rand() & 1)
    f[x] = y;
  else
    f[y] = x;
}
void tarjan(int x, int pre) {
  dfn[x] = low[x] = ++idx;
  for (__typeof(p[x].begin()) it = p[x].begin(); it != p[x].end(); ++it) {
    if (pre == (*it).second) continue;
    int y = (*it).first;
    if (dfn[y]) {
      low[x] = min(low[x], dfn[y]);
    } else {
      tarjan(y, (*it).second);
      low[x] = min(low[x], low[y]);
      if (dfn[x] < low[y]) res[(*it).second] = 1;
    }
  }
}
int main() {
  srand(time(NULL));
  cin >> n >> m;
  int x, y, d;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d%d", &x, &y, &d);
    msg[i].idx = i;
    msg[i].d = d;
    msg[i].x = x;
    msg[i].y = y;
  }
  sort(msg, msg + m);
  init();
  int i = 0, j = 0;
  while (i < m) {
    set<int> st;
    for (; j < m && msg[j].d == msg[i].d; ++j)
      if ((x = root(msg[j].x)) != (y = root(msg[j].y))) {
        st.insert(x);
        st.insert(y);
        p[x].push_back(make_pair(y, msg[j].idx));
        p[y].push_back(make_pair(x, msg[j].idx));
        res[msg[j].idx] = 2;
      }
    for (; i < j; ++i) unions(msg[i].x, msg[i].y);
    for (__typeof(st.begin()) it = st.begin(); it != st.end(); ++it)
      if (!dfn[*it]) {
        tarjan(*it, -1);
      }
    for (__typeof(st.begin()) it = st.begin(); it != st.end(); ++it) {
      p[*it].clear();
      dfn[*it] = low[*it] = 0;
    }
  }
  for (int i = 0; i < m; ++i) puts(inf[res[i]]);
  return 0;
}
