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;
bool debug = false;
long long k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
struct Edge {
  int u, v, id, w;
  bool operator<(const Edge &b) const { return w < b.w; }
} e[100005];
int ans[100005], f[100005], cnt1 = 1, eg[100005 * 2], head[100005 * 2],
                            nxt[100005 * 2], id[100005 * 2];
bool vis[100005];
vector<int> g;
int ff(int x) { return x == f[x] ? x : f[x] = ff(f[x]); }
void add(int u, int v, int i) {
  eg[++cnt1] = v;
  nxt[cnt1] = head[u];
  head[u] = cnt1;
  id[cnt1] = i;
}
int dfn[100005], low[100005], sta[100005], T, cnt, belong[100005], color;
void tarjan(int u, int p) {
  dfn[u] = low[u] = ++T;
  sta[++cnt] = u;
  for (int i = head[u]; i; i = nxt[i]) {
    int child = eg[i];
    if (i != (p ^ 1)) {
      if (!dfn[child]) {
        tarjan(child, i);
        low[u] = min(low[u], low[child]);
      } else if (!belong[child]) {
        low[u] = min(low[u], dfn[child]);
      }
    }
  }
  if (dfn[u] == low[u]) {
    color++;
    while (cnt) {
      int v = sta[cnt--];
      belong[v] = color;
      if (v == u) break;
    }
  }
}
void cal() {
  sort(e + 1, e + 1 + m);
  int sum = n;
  for (int i = 1; i <= m + 1; i++) {
    if (i != 1 && e[i].w != e[i - 1].w) {
      cout << "";
      for (int x : g)
        if (!dfn[x]) tarjan(x, 0);
      for (int u : g) {
        for (int i = head[u]; i; i = nxt[i]) {
          int v = eg[i];
          if (belong[u] != belong[v])
            ans[id[i]] = 1;
          else
            ans[id[i]] = 2;
          int u1 = ff(u), v1 = ff(v);
          if (u1 == v1) continue;
          f[u1] = v1;
          sum--;
        }
      }
      cnt1 = 1;
      T = 0;
      color = 0;
      for (int x : g) {
        vis[x] = belong[x] = dfn[x] = low[x] = head[x] = 0;
      }
      g.clear();
    }
    if (i == m + 1) break;
    int u = ff(e[i].u), v = ff(e[i].v);
    if (u == v) continue;
    if (!vis[u]) {
      g.push_back(u);
      vis[u] = 1;
    }
    if (!vis[v]) {
      g.push_back(v);
      vis[v] = 1;
    }
    add(u, v, e[i].id);
    add(v, u, e[i].id);
    if (sum == 1) break;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
    e[i].id = i;
  }
  cal();
  for (int i = 1; i <= m; i++) {
    if (ans[i] == 1)
      puts("any");
    else if (ans[i] == 2)
      puts("at least one");
    else
      puts("none");
  }
  return 0;
}
