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 maxn = 100000 + 10;
const int maxm = 100000 + 10;
struct Edge {
  int from, to, weight, id;
  bool operator<(const Edge& rhs) const { return weight < rhs.weight; }
};
struct E {
  int to, next, id;
};
int pa[maxn], ans[maxn];
Edge e[maxm];
int findset(int x) { return pa[x] == x ? x : (pa[x] = findset(pa[x])); }
struct Bridge {
  int m;
  int head[maxn];
  E edges[maxm * 2];
  int pre[maxn], dfs_clock;
  void init(int n) {
    m = dfs_clock = 0;
    for (int i = 1; i <= n; i++) {
      head[i] = -1;
      pre[i] = 0;
    }
  }
  void AddEdge(int from, int to, int id) {
    edges[m].next = head[from];
    edges[m].to = to;
    edges[m].id = id;
    head[from] = m++;
  }
  int dfs(int u, int fa) {
    int lowu = pre[u] = ++dfs_clock;
    for (int i = head[u]; i != -1; i = edges[i].next) {
      int v = edges[i].to;
      if (!pre[v]) {
        int lowv = dfs(v, edges[i].id);
        lowu = min(lowu, lowv);
        if (lowv > pre[u]) {
          ans[edges[i].id] = 2;
        }
      } else if (edges[i].id != fa) {
        lowu = min(lowu, pre[v]);
      }
    }
    return lowu;
  }
};
Bridge br;
int N, M;
char res[][20] = {"none", "at least one", "any"};
void init() {
  for (int i = 1; i <= N; i++) {
    pa[i] = i;
  }
  for (int i = 0; i < M; i++) {
    ans[i] = 0;
  }
}
void Pre() {
  init();
  for (int i = 0; i < M; i++) {
    scanf("%d%d%d", &e[i].from, &e[i].to, &e[i].weight);
    e[i].id = i;
  }
  sort(e, e + M);
  br.init(N);
}
int main() {
  while (scanf("%d%d", &N, &M) != EOF) {
    Pre();
    for (int i = 0; i < M;) {
      int start = i;
      while (i < M && e[i].weight == e[start].weight) i++;
      for (int j = start; j < i; j++) {
        int u = e[j].from, v = e[j].to;
        int x = findset(u), y = findset(v);
        if (x != y) {
          br.AddEdge(x, y, e[j].id);
          br.AddEdge(y, x, e[j].id);
          ans[e[j].id] = 1;
        }
      }
      for (int j = start; j < i; j++) {
        int x = findset(e[j].from), y = findset(e[j].to);
        if (x != y && !br.pre[x]) br.dfs(x, -1);
      }
      for (int j = start; j < i; j++) {
        int u = e[j].from, v = e[j].to;
        int x = findset(u), y = findset(v);
        if (x != y) {
          pa[y] = x;
          br.head[x] = br.head[y] = -1;
          br.pre[x] = br.pre[y] = 0;
        }
      }
      br.m = br.dfs_clock = 0;
    }
    for (int i = 0; i < M; i++) {
      printf("%s\n", res[ans[i]]);
    }
  }
  return 0;
}
