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 = 1e5 + 7, M = 3e5 + 7;
int read() {
  int s = 0, w = 1;
  char c;
  while (c = getchar(), c > '9' || c < '0')
    if (c == '-') w = -1;
  while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
  return s * w;
}
struct edge {
  int u, v, w, id;
  bool operator<(edge a) const { return w < a.w; }
} e[M];
struct Edge {
  int t, next, id;
} E[M << 1];
int fa[N], ans[M], head[N], vis[N];
int n, m, cnt = 1;
vector<int> g;
void add(int u, int t, int id) {
  E[++cnt].t = t;
  E[cnt].next = head[u];
  E[cnt].id = id;
  head[u] = cnt;
}
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int tp, T, Scc;
int scc[N], st[N], low[N], dfn[N];
void Tarjan(int u, int fa) {
  st[++tp] = u;
  low[u] = dfn[u] = ++T;
  for (int i = head[u]; i; i = E[i].next) {
    int v = E[i].t;
    if (i == (fa ^ 1)) continue;
    if (!dfn[v]) {
      Tarjan(v, i);
      low[u] = min(low[u], low[v]);
    } else if (!scc[v]) {
      low[u] = min(low[u], dfn[v]);
    }
  }
  if (low[u] == dfn[u]) {
    Scc++;
    while (tp) {
      int v = st[tp--];
      scc[v] = Scc;
      if (v == u) break;
    }
  }
}
void solve() {
  sort(e + 1, e + m + 1);
  int sum = n;
  for (int i = 1; i <= m + 1; i++) {
    if (i != 1 && e[i].w != e[i - 1].w) {
      for (int j = 0; j < (int)g.size(); j++)
        if (!dfn[g[j]]) Tarjan(g[j], 0);
      for (int j = 0; j < (int)g.size(); j++) {
        int u = g[j];
        for (int k = head[u]; k; k = E[k].next) {
          if (scc[u] != scc[E[k].t])
            ans[E[k].id] = 1;
          else
            ans[E[k].id] = 2;
          int u1 = find(u), v1 = find(E[k].t);
          if (u1 == v1) continue;
          fa[v1] = u1;
          sum--;
        }
      }
      cnt = 1;
      T = 0;
      Scc = 0;
      for (int j = 0; j < (int)g.size(); j++)
        vis[g[j]] = scc[g[j]] = dfn[g[j]] = low[g[j]] = head[g[j]] = 0;
      g.clear();
    }
    if (i == m + 1) break;
    int u = find(e[i].u), v = find(e[i].v);
    if (u == v) continue;
    if (!vis[u]) g.push_back(u), vis[u] = 1;
    if (!vis[v]) g.push_back(v), vis[v] = 1;
    add(u, v, e[i].id);
    add(v, u, e[i].id);
    if (sum == 1) break;
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= m; i++) {
    e[i].u = read(), e[i].v = read(), e[i].w = read();
    e[i].id = i;
  }
  solve();
  for (int i = 1; i <= m; i++) {
    if (ans[i] == 1)
      puts("any");
    else if (ans[i] == 2)
      puts("at least one");
    else
      puts("none");
  }
}
