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 = 210000;
const int INF = 123456789;
struct node {
  int id, u, v, w, lazy;
  node() { lazy = -1; }
};
node edge[MAXN];
struct node2 {
  int v, w, pos;
  node2 *next, *op;
  bool vis;
};
node2 sonedge[MAXN];
node2 *ecnt = &sonedge[0];
node2 *adj[MAXN];
void addedge(int u, int v, int w, int j) {
  node2 *p = ++ecnt;
  p->v = v;
  p->w = w;
  p->pos = j;
  p->op = ecnt + 1;
  p->vis = false;
  p->next = adj[u];
  adj[u] = p;
  p = ++ecnt;
  p->v = u;
  p->w = w;
  p->pos = j;
  p->op = ecnt - 1;
  p->vis = false;
  p->next = adj[v];
  adj[v] = p;
}
int dfn[MAXN], low[MAXN], dcnt;
int n, m;
int s[MAXN], t[MAXN], cnt;
int fa[MAXN];
bool abk[MAXN];
void read() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
    edge[i].id = i;
  }
}
bool cmp(node a, node b) {
  if (a.w < b.w)
    return true;
  else
    return false;
}
bool cmp2(node a, node b) {
  if (a.id < b.id)
    return true;
  else
    return false;
}
int find(int p) {
  if (fa[p] == 0) return p;
  return fa[p] = find(fa[p]);
}
void dfs(int u) {
  dfn[u] = low[u] = ++dcnt;
  for (node2 *p = adj[u]; p != NULL; p = p->next)
    if (p->vis == false) {
      p->vis = true;
      p->op->vis = true;
      if (dfn[p->v] == 0) {
        dfs(p->v);
        low[u] = min(low[u], low[p->v]);
        if (low[p->v] > dfn[u]) abk[p->pos] = true;
      } else
        low[u] = min(low[u], low[p->v]);
    }
}
void solve() {
  int i, j;
  int x, y;
  sort(edge + 1, edge + m + 1, cmp);
  edge[0].w = -INF;
  for (i = 1; i <= m; i++)
    if (edge[i].w != edge[i - 1].w) {
      t[cnt] = i - 1;
      s[++cnt] = i;
    }
  t[cnt] = m;
  for (i = 1; i <= cnt; i++) {
    ecnt = &sonedge[0];
    for (j = s[i]; j <= t[i]; j++) {
      x = find(edge[j].u);
      y = find(edge[j].v);
      if (x == y)
        edge[j].lazy = 0;
      else {
        edge[j].lazy = 1;
        addedge(x, y, edge[j].w, edge[j].id);
      }
    }
    for (j = s[i]; j <= t[i]; j++) {
      x = find(edge[j].u);
      if (dfn[x] == 0) dfs(x);
    }
    for (j = s[i]; j <= t[i]; j++)
      if (abk[edge[j].id] == true) edge[j].lazy = 2;
    for (j = s[i]; j <= t[i]; j++) {
      x = find(edge[j].u);
      y = find(edge[j].v);
      if (x != y) fa[y] = x;
      adj[x] = adj[y] = NULL;
      dfn[edge[j].u] = dfn[edge[j].v] = 0;
      low[edge[j].u] = low[edge[j].v] = 0;
      dfn[x] = dfn[y] = 0;
      low[x] = low[y] = 0;
    }
  }
  sort(edge + 1, edge + m + 1, cmp2);
  for (i = 1; i <= m; i++) {
    if (edge[i].lazy == 2)
      printf("any\n");
    else if (edge[i].lazy == 1)
      printf("at least one\n");
    else if (edge[i].lazy == 0)
      printf("none\n");
  }
}
int main() {
  read();
  solve();
  return 0;
}
