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 = 100007;
const int M = 100007;
struct Edge {
  int v, id;
  Edge *nxt;
} edge[2 * M];
Edge *ecnt, *NIL;
Edge *head[N];
struct Data {
  int u, v, w, id;
} e[M];
bool cmp(Data x, Data y) { return x.w < y.w; }
int n, m, pcnt, ans[M];
int fa[N], dfn[N];
int Find(int x) {
  if (fa[x] == x) return x;
  return fa[x] = Find(fa[x]);
}
void Link(int u, int v) {
  int fu = Find(u);
  int fv = Find(v);
  if (fu != fv) {
    head[fu] = head[fv] = NIL;
    dfn[fu] = dfn[fv] = 0;
    fa[fu] = fv;
  }
}
void AddEdge(int u, int v, int id) {
  ecnt->v = v;
  ecnt->id = id;
  ecnt->nxt = head[u];
  head[u] = ecnt++;
}
int Dfs(int u, int pre) {
  int low = dfn[u] = ++pcnt;
  for (Edge *g = head[u]; g != NIL; g = g->nxt) {
    if (!dfn[g->v]) {
      int tmp = Dfs(g->v, g->id);
      low = min(low, tmp);
      if (tmp > dfn[u]) ans[g->id] = 2;
    } else if (dfn[g->v] < dfn[u] && g->id != pre)
      low = min(low, dfn[g->v]);
  }
  return low;
}
int main() {
  NIL = &edge[0];
  scanf("%d %d", &n, &m);
  for (int i = 1; i < N; i++) fa[i] = i, head[i] = NIL;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
    e[i].id = i;
  }
  sort(e + 1, e + m + 1, cmp);
  for (int i = 1, j = 0; j <= m; i = j) {
    ecnt = &edge[1];
    j = i + 1;
    while (j <= m && e[j].w == e[i].w) j++;
    for (int k = i; k < j; k++) {
      int fu = Find(e[k].u);
      int fv = Find(e[k].v);
      if (fu != fv) {
        AddEdge(fu, fv, e[k].id);
        AddEdge(fv, fu, e[k].id);
        ans[e[k].id] = 1;
      }
    }
    for (int k = i; k < j; k++) {
      int fu = Find(e[k].u);
      int fv = Find(e[k].v);
      if (fu != fv && !dfn[fu]) Dfs(fu, -1);
    }
    for (int k = i; k < j; k++) Link(e[k].u, e[k].v);
  }
  for (int i = 1; i <= m; i++)
    if (ans[i] == 2)
      puts("any");
    else if (!ans[i])
      puts("none");
    else
      puts("at least one");
  return 0;
}
