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;
int n, m, first[100010] = {0}, tot = 0, pre[100010] = {0}, ans[100010] = {0};
int dep[100010] = {0}, fa[100010], preid[100010], prew[100010], fak[100010];
struct kruskaledge {
  int id, a, b, w;
  bool type;
} e[100010];
struct edge {
  int id, v, w, next;
} ed[200010];
bool cmp(kruskaledge a, kruskaledge b) { return a.w < b.w; }
bool cmp2(kruskaledge a, kruskaledge b) {
  if (a.w == b.w)
    return a.type < b.type;
  else
    return a.w < b.w;
}
void insert(int a, int b, int id, int w) {
  ed[++tot].v = b;
  ed[tot].next = first[a];
  ed[tot].id = id;
  ed[tot].w = w;
  first[a] = tot;
}
int find(int *fa, int x) {
  int r = x, i = x, j;
  while (fa[r] != r) r = fa[r];
  while (i != r) j = fa[i], fa[i] = r, i = j;
  return r;
}
void merge(int *fa, int x, int y) {
  int fx = find(fa, x), fy = find(fa, y);
  fa[fx] = fy;
}
void kruskal() {
  sort(e + 1, e + m + 1, cmp);
  for (int i = 1; i <= n; i++) fa[i] = i;
  for (int i = 1; i <= m; i++) {
    if (find(fa, e[i].a) != find(fa, e[i].b)) {
      e[i].type = 1;
      merge(fa, e[i].a, e[i].b);
      insert(e[i].a, e[i].b, e[i].id, e[i].w);
      insert(e[i].b, e[i].a, e[i].id, e[i].w);
    } else
      e[i].type = 0;
  }
}
void dfs(int v) {
  for (int i = first[v]; i; i = ed[i].next)
    if (ed[i].v != pre[v]) {
      pre[ed[i].v] = v;
      dep[ed[i].v] = dep[v] + 1;
      preid[ed[i].v] = ed[i].id;
      prew[ed[i].v] = ed[i].w;
      dfs(ed[i].v);
    }
}
void up(int &x, int w) {
  if (prew[x] == w) ans[preid[x]] = 1;
  int nxt = find(fak, pre[x]);
  merge(fak, x, nxt);
  x = nxt;
}
void solve() {
  sort(e + 1, e + m + 1, cmp2);
  for (int i = 1; i <= n; i++) fa[i] = fak[i] = i;
  for (int i = 1; i <= m; i++) {
    if (e[i].type)
      merge(fa, e[i].a, e[i].b);
    else {
      if (find(fa, e[i].a) != find(fa, e[i].b)) {
        ans[e[i].id] = 1;
        int x = e[i].a, y = e[i].b;
        while (x != y) {
          if (dep[x] == dep[y]) {
            up(x, e[i].w);
            up(y, e[i].w);
          } else {
            if (dep[x] < dep[y]) swap(x, y);
            up(x, e[i].w);
          }
        }
      } else
        ans[e[i].id] = 2;
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    e[i].id = i;
    scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].w);
  }
  kruskal();
  dep[1] = 1;
  dfs(1);
  solve();
  for (int i = 1; i <= m; i++) {
    if (ans[i] == 0) printf("any\n");
    if (ans[i] == 1) printf("at least one\n");
    if (ans[i] == 2) printf("none\n");
  }
  return 0;
}
