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 = (int)1e5 + 10;
int n, m, ans[M], head[2 * M], f[M], cnt, low[M], pre[M], dfs_clock, vis[M];
struct node1 {
  int to, w, next;
} e[2 * M];
struct node2 {
  int a, b, w, pos;
  bool operator<(const node2& y) const { return w < y.w; }
} a[M];
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void add(int from, int to, int w) {
  e[cnt].to = to;
  e[cnt].w = w;
  e[cnt].next = head[from];
  head[from] = cnt++;
}
void merge(int x, int y) {
  int fx = find(x), fy = find(y);
  if (fx != fy) {
    pre[fx] = pre[fy] = 0;
    head[fx] = head[fy] = -1;
    low[fx] = low[fy] = 0;
    vis[fx] = vis[fy] = 0;
    f[fx] = fy;
  }
}
int dfs(int x, int p) {
  int lowu = pre[x] = ++dfs_clock;
  vis[x] = 1;
  for (int i = head[x]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (!vis[v]) {
      int lowv = dfs(v, e[i].w);
      lowu = min(lowu, lowv);
      if (lowv > pre[x]) ans[e[i].w] = 1;
    } else if (e[i].w != p)
      lowu = min(lowu, pre[v]);
  }
  return low[x] = lowu;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) f[i] = i;
  memset(head, -1, sizeof(head));
  for (int i = 0; i < m; i++)
    scanf("%d %d %d", &a[i].a, &a[i].b, &a[i].w), a[i].pos = i;
  sort(a, a + m);
  for (int i = 0; i < m; i++) {
    int t = i;
    while (t + 1 < m && a[t].w == a[t + 1].w) t++;
    for (int j = i; j <= t; j++) {
      int fx = find(a[j].a), fy = find(a[j].b);
      if (fx != fy) {
        ans[a[j].pos] = 2;
        add(fx, fy, a[j].pos);
        add(fy, fx, a[j].pos);
      }
    }
    for (int j = i; j <= t; j++) {
      int fx = find(a[j].a), fy = find(a[j].b);
      if (fx != fy && !vis[fx]) dfs(fx, -1);
      if (fx != fy && !vis[fy]) dfs(fy, -1);
    }
    for (int j = i; j <= t; j++) merge(a[j].a, a[j].b);
    i = t;
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == 0)
      printf("none\n");
    else if (ans[i] == 1)
      printf("any\n");
    else
      printf("at least one\n");
  }
  return 0;
}
