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 = 100005;
int n, m;
struct Edge {
  int x, y, val, id, ans;
  bool operator<(const Edge& a) const { return val < a.val; }
} e[MAXN];
struct E {
  int v, next, id;
} edge[MAXN * 2];
int fa[MAXN];
int low[MAXN], dep[MAXN];
int head[MAXN], len;
int mark[MAXN];
int com[MAXN];
char res[][20] = {"none", "any", "at least one"};
void add_edge(int u, int v, int id) {
  edge[len].v = v;
  edge[len].id = id;
  edge[len].next = head[u];
  head[u] = len++;
}
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
bool cmp(Edge a, Edge b) { return a.id < b.id; }
void dfs(int u, int father, int depth) {
  int i;
  mark[u] = 1;
  low[u] = dep[u] = depth;
  for (i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (mark[v] == 1 && edge[i].id != father && dep[v] < low[u])
      low[u] = dep[v];
    else if (mark[v] == 0) {
      dfs(v, edge[i].id, depth + 1);
      if (low[v] < low[u]) low[u] = low[v];
      if (low[v] > dep[u]) e[edge[i].id].ans = 1;
    }
  }
  mark[u] = 2;
}
int main() {
  int i, j, x, y;
  scanf("%d%d", &n, &m);
  for (i = 0; i <= n; i++) fa[i] = i;
  for (i = 0; i < m; i++) {
    scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].val);
    e[i].id = i;
    e[i].ans = 0;
  }
  sort(e, e + m);
  memset(head, -1, sizeof(head));
  len = 0;
  for (i = 0; i < m;) {
    int s = i;
    while (i < m && e[i].val == e[s].val) i++;
    for (j = s; j < i; j++) {
      x = find(e[j].x);
      y = find(e[j].y);
      if (x == y) continue;
      e[j].ans = 2;
      add_edge(x, y, j);
      add_edge(y, x, j);
      com[j] = x;
    }
    for (j = s; j < i; j++) {
      if (com[j] && !mark[com[j]]) dfs(com[j], 0, 1);
    }
    for (j = s; j < i; j++) {
      x = find(e[j].x);
      y = find(e[j].y);
      head[x] = -1;
      head[y] = -1;
      mark[x] = 0;
      mark[y] = 0;
      if (x == y) continue;
      fa[x] = y;
    }
    len = 0;
  }
  sort(e, e + m, cmp);
  for (i = 0; i < m; i++) printf("%s\n", res[e[i].ans]);
  return 0;
}
