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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long inf = 1ll << 62;
inline long long read() {
  register long long x = 0, f = 1;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
void writeln(long long x) {
  write(x);
  puts("");
}
const int maxn = 100233;
struct Edge {
  int x, y, z, id;
  bool operator<(const Edge &w) const { return z < w.z; }
} edge[maxn];
vector<pair<int, int> > V[maxn];
int fa[maxn], ans[maxn], pre[maxn], n, m;
int dfsclk = 0;
int dfs(int u, int fa) {
  int lowu = pre[u] = ++dfsclk;
  for (unsigned i = 0; i < V[u].size(); ++i) {
    int v = V[u][i].first;
    if (!pre[v]) {
      int lowv = dfs(v, V[u][i].second);
      lowu = ((lowu) < (lowv) ? (lowu) : (lowv));
      if (lowv > pre[u]) ans[V[u][i].second] = 1;
    } else {
      if (pre[v] < pre[u] && V[u][i].second != fa)
        lowu = ((lowu) < (pre[v]) ? (lowu) : (pre[v]));
    }
  }
  return lowu;
}
inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void join(int x, int y) {
  x = find(x), y = find(y);
  if (x != y) {
    V[x].clear(), V[y].clear();
    pre[x] = pre[y] = 0;
    fa[x] = y;
  }
}
int main() {
  dfsclk = 0;
  n = read(), m = read();
  for (register int i = (1); i <= int(n); ++i) V[i].clear();
  for (register int i = (1); i <= int(m); ++i) {
    edge[i].x = read();
    edge[i].y = read();
    edge[i].z = read();
    edge[i].id = i;
  }
  memset(ans, 0, sizeof(ans));
  memset(pre, 0, sizeof(pre));
  for (register int i = (1); i <= int(n); ++i) fa[i] = i;
  sort(edge + 1, edge + 1 + m);
  for (int i = 1, j; i <= m; i = j + 1) {
    for (j = i; j + 1 <= m && edge[j + 1].z == edge[i].z; ++j)
      ;
    int cnt = 0;
    for (register int k = (i); k <= int(j); ++k) {
      int x = edge[k].x, y = edge[k].y;
      if (find(x) != find(y)) {
        V[find(x)].push_back(make_pair(find(y), edge[k].id));
        V[find(y)].push_back(make_pair(find(x), edge[k].id));
        ans[edge[k].id] = 2;
      }
    }
    for (register int k = (i); k <= int(j); ++k) {
      int x = edge[k].x, y = edge[k].y;
      if (find(x) != find(y) && !pre[find(x)]) {
        dfs(find(x), -1);
      }
    }
    for (register int k = (i); k <= int(j); ++k) join(edge[k].x, edge[k].y);
  }
  for (register int i = (1); i <= int(m); ++i) {
    if (ans[i] == 0)
      puts("none");
    else if (ans[i] == 1)
      puts("any");
    else
      puts("at least one");
  }
  return 0;
}
