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;
struct drum {
  int x, y, c, id;
  bool all = false, none = false;
  bool operator<(drum b) const { return c < b.c; }
};
struct nodS {
  int par, d, dp;
  bool viz;
  vector<int> v;
};
bool compid(drum a, drum b) { return a.id < b.id; }
int n, m, i, j, dr, t;
drum e[100005];
nodS nod[100005];
int dsu[100005];
map<pair<int, int>, int> ma;
stack<int> s;
int dsu_grup(int x) {
  if (dsu[x] == x) return x;
  dsu[x] = dsu_grup(dsu[x]);
  return dsu[x];
}
void dsu_merge(int x, int y) {
  x = dsu_grup(x);
  y = dsu_grup(y);
  if (x != y) dsu[y] = x;
}
void refr(int x) {
  if (nod[x].viz) {
    nod[x].viz = false;
    nod[x].dp = 0;
    nod[x].v.clear();
  }
}
void dfs(int p, int par) {
  nod[p].viz = true;
  nod[p].d = nod[par].d + 1;
  nod[p].par = par;
  for (int it : nod[p].v) {
    if (it == par) continue;
    if (nod[it].viz) {
      if (nod[it].d < nod[p].d) {
        nod[p].dp++;
        nod[it].dp--;
      }
    } else {
      dfs(it, p);
      nod[p].dp += nod[it].dp;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    dsu[i] = i;
  }
  for (i = 1; i <= m; i++) {
    cin >> e[i].x >> e[i].y >> e[i].c;
    e[i].id = i;
  }
  sort(e + 1, e + m + 1);
  for (i = 1; i <= m; i++) {
    ma.clear();
    t = 0;
    for (j = i; j <= m; j++) {
      if (e[j].c == e[i].c)
        dr = j;
      else
        break;
    }
    for (j = i; j <= dr; j++) {
      e[j].x = dsu_grup(e[j].x);
      e[j].y = dsu_grup(e[j].y);
      if (e[j].x == e[j].y) {
        e[j].none = true;
        continue;
      }
      if (e[j].x > e[j].y) swap(e[j].x, e[j].y);
      ma[{e[j].x, e[j].y}]++;
      refr(e[j].x);
      refr(e[j].y);
      nod[e[j].x].v.push_back(e[j].y);
      nod[e[j].y].v.push_back(e[j].x);
    }
    for (j = i; j <= dr; j++) {
      if (e[j].none) continue;
      if (!nod[e[j].x].viz) dfs(e[j].x, 0);
      if (nod[e[j].x].d < nod[e[j].y].d) swap(e[j].x, e[j].y);
      if (nod[e[j].x].par == e[j].y && nod[e[j].x].dp == 0) e[j].all = true;
      if (e[j].x > e[j].y) swap(e[j].x, e[j].y);
      if (ma[{e[j].x, e[j].y}] > 1) e[j].all = false;
    }
    for (j = i; j <= dr; j++) {
      if (!e[j].none) dsu_merge(e[j].x, e[j].y);
    }
    i = dr;
  }
  sort(e + 1, e + m + 1, compid);
  for (i = 1; i <= m; i++) {
    if (e[i].none) {
      cout << "none\n";
      continue;
    }
    if (e[i].all) {
      cout << "any\n";
    } else {
      cout << "at least one\n";
    }
  }
  return 0;
}
