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 M = 200005;
int ans[M], low[M], dfn[M], head[M], n, m, fa[M], tot = 0, timestamp = 0;
struct node {
  int to, nex, id;
} e[M * 2];
struct edge {
  int a, b, c, id;
} ed[M];
inline void rd(int &res) {
  res = 0;
  char c;
  while (c = getchar(), c < 48)
    ;
  do res = (res << 3) + (res << 1) + (c ^ 48);
  while (c = getchar(), c > 47);
}
void print(int x) {
  if (x == 0) return;
  print(x / 10);
  putchar((x % 10) ^ 48);
}
bool cmp(edge a, edge b) { return a.c < b.c; }
int get(int x) {
  if (fa[x] != x) fa[x] = get(fa[x]);
  return fa[x];
}
void dfs(int x, int id) {
  dfn[x] = low[x] = ++timestamp;
  for (int i = head[x]; ~i; i = e[i].nex) {
    int to = e[i].to, eid = e[i].id;
    if (dfn[to] == 0) {
      dfs(to, i);
      low[x] = min(low[x], low[to]);
      if (low[to] > dfn[x]) ans[eid] = 1;
    } else if (dfn[to] < dfn[x] && i != (id ^ 1))
      low[x] = min(low[x], low[to]);
  }
}
void add_edge(int a, int b, int id) {
  e[tot] = (node){b, head[a], id};
  head[a] = tot++;
}
void get_union(int a, int b) {
  a = get(a);
  b = get(b);
  if (a != b) fa[a] = b;
}
int main() {
  int i, j, k, a, b, c, en;
  rd(n);
  rd(m);
  for (i = 1; i <= m; i++) {
    rd(a);
    rd(b);
    rd(c);
    ed[i] = (edge){a, b, c, i};
  }
  sort(ed + 1, ed + 1 + m, cmp);
  for (i = 1; i <= n; i++) fa[i] = i, head[i] = -1;
  for (i = 1; i <= m; i++) {
    for (j = i; j <= m; j++) {
      if (ed[j + 1].c != ed[i].c) break;
    }
    en = j;
    for (j = i; j <= en; j++) {
      ed[j].a = a = get(ed[j].a);
      ed[j].b = b = get(ed[j].b);
      dfn[a] = 0;
      dfn[b] = 0;
      low[a] = 0;
      low[b] = 0;
      head[a] = -1;
      head[b] = -1;
    }
    for (j = i; j <= en; j++) {
      a = ed[j].a;
      b = ed[j].b;
      if (a == b) {
        ans[ed[j].id] = 2;
        continue;
      }
      add_edge(a, b, ed[j].id);
      add_edge(b, a, ed[j].id);
    }
    for (j = i; j <= en; j++) {
      if (!dfn[ed[j].a]) dfs(ed[j].a, -1);
      if (!dfn[ed[j].b]) dfs(ed[j].b, -1);
    }
    for (j = i; j <= en; j++) {
      get_union(ed[j].a, ed[j].b);
    }
    i = en;
  }
  for (i = 1; i <= m; i++)
    if (ans[i] == 2)
      puts("none");
    else if (ans[i] == 1)
      puts("any");
    else
      puts("at least one");
  return 0;
}
