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 N = 100005, M = N * 2;
struct Gragh {
  int cnt, y[M], nxt[M], fst[N], z[M];
  void clear() {
    cnt = 0;
    memset(fst, 0, sizeof fst);
  }
  void add(int a, int b, int c) {
    y[++cnt] = b, nxt[cnt] = fst[a], fst[a] = cnt;
    z[cnt] = c;
  }
} g;
struct Edge {
  int x, y, z, id;
} e[N];
bool cmp(Edge a, Edge b) { return a.z < b.z; }
int ffa[N];
int n, m;
int getf(int x) { return ffa[x] == x ? x : ffa[x] = getf(ffa[x]); }
int depth[N], fa[N][20], son[N], size[N], p[N], top[N], ap[N];
int tax[N * 10], ans[N], f[N], Min[N][20];
vector<int> c[N];
void dfs(int x, int pre, int d) {
  depth[x] = d, fa[x][0] = pre, size[x] = 0, son[x] = -1;
  for (int i = 1; i < 20; i++) {
    fa[x][i] = fa[fa[x][i - 1]][i - 1];
    int a = Min[x][i - 1], b = Min[fa[x][i - 1]][i - 1];
    Min[x][i] = e[a].z > e[b].z ? a : b;
  }
  for (int i = g.fst[x]; i; i = g.nxt[i])
    if (g.y[i] != pre) {
      Min[g.y[i]][0] = g.z[i];
      dfs(g.y[i], x, d + 1);
      size[x] += size[g.y[i]];
      if (son[x] == -1 || size[g.y[i]] > size[son[x]]) son[x] = g.y[i];
    }
}
int cnp = 0;
void Get_Top(int x, int Top) {
  top[x] = Top;
  ap[p[x] = ++cnp] = x;
  if (son[x] == -1) return;
  Get_Top(son[x], Top);
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    int y = g.y[i];
    if (y != fa[x][0] && y != son[x]) Get_Top(y, y);
  }
}
int LCA(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  for (int i = 19; i >= 0; i--)
    if (depth[x] - (1 << i) >= depth[y]) x = fa[x][i];
  if (x == y) return x;
  for (int i = 19; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
int ask(int x, int d) {
  int ans = 0;
  for (int i = 19; i >= 0; i--)
    if (d >= (1 << i)) {
      d -= 1 << i;
      ans = e[ans].z > e[Min[x][i]].z ? ans : Min[x][i];
      x = fa[x][i];
    }
  return ans;
}
void Tupdate(int x, int lca, int i) {
  while (top[x] != top[lca]) {
    int f = top[x];
    c[p[f]].push_back(i);
    c[p[x]].push_back(-i);
    x = fa[f][0];
    if (e[Min[f][0]].z == e[i].z) ans[Min[f][0]] = 1;
  }
  c[p[lca]].push_back(i);
  c[p[x]].push_back(-i);
}
int rans[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++)
    scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].z), e[i].id = i;
  sort(e + 1, e + m + 1, cmp);
  for (int i = 1; i <= n; i++) ffa[i] = i;
  g.clear();
  for (int i = 1; i <= m; i++) {
    int x = e[i].x, y = e[i].y, z = e[i].z;
    int px = getf(x), py = getf(y);
    if (px == py) continue;
    f[i] = 1;
    ans[i] = 2;
    ffa[px] = py;
    g.add(x, y, i);
    g.add(y, x, i);
  }
  dfs(1, 0, 0);
  Get_Top(1, 0);
  for (int i = 1; i <= m; i++) {
    if (f[i]) continue;
    int x = e[i].x, y = e[i].y, lca = LCA(x, y);
    int aa = ask(x, depth[x] - depth[lca]);
    int bb = ask(y, depth[y] - depth[lca]);
    ans[i] = (bool)(max(e[aa].z, e[bb].z) == e[i].z);
    Tupdate(x, lca, i);
    Tupdate(y, lca, i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < c[i].size(); j++)
      if (c[i][j] < 0)
        tax[e[-c[i][j]].z]--;
      else
        tax[e[c[i][j]].z]++;
    if (tax[e[Min[ap[i + 1]][0]].z] > 0) ans[Min[ap[i + 1]][0]] = 1;
  }
  for (int i = 1; i <= m; i++) rans[e[i].id] = ans[i];
  for (int i = 1; i <= m; i++) {
    if (rans[i] == 0) puts("none");
    if (rans[i] == 1) puts("at least one");
    if (rans[i] == 2) puts("any");
  }
  return 0;
}
