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;
template <class X>
inline void cmax(X& _, X __) {
  if (_ < __) _ = __;
}
template <class X>
inline void cmin(X& _, X __) {
  if (_ > __) _ = __;
}
template <class X>
inline int read(X& x) {
  int c = getchar(), f = (x = 0);
  while (~c && (c < 48 || c > 57)) {
    if (c == '-') {
      f = 1;
    }
    c = getchar();
  }
  if (!~c) return 0;
  while (c > 47 && c < 58) {
    x = x * 10 + (c ^ 48), c = getchar();
  }
  if (f) {
    x = -x;
  }
  return 1;
}
template <class X, class Y>
int read(X& x, Y& y) {
  return read(x) && read(y);
}
template <class X, class Y, class Z>
int read(X& x, Y& y, Z& z) {
  return read(x) && read(y, z);
}
const int maxn = 1e5 + 10, INF = 0x3f3f3f3f;
const long long LINF = (long long)INF << 32 | INF, mod = 1e9 + 7;
struct Edge {
  int u, v, w, id;
  Edge(int a, int b, int c, int d) : u(a), v(b), w(c), id(d) {}
  bool operator<(const Edge& T) const { return w < T.w; }
};
struct Node {
  int v, id;
  Node(int a, int b) : v(a), id(b) {}
};
vector<Edge> edge;
vector<Node> a[maxn];
int fa[maxn], ans[maxn], dfn[maxn], low[maxn], tot;
void Tarjan(int u, int pre) {
  dfn[u] = low[u] = ++tot;
  for (Node& x : a[u])
    if (x.id != pre) {
      if (!dfn[x.v]) {
        Tarjan(x.v, x.id);
        cmin(low[u], low[x.v]);
        if (low[x.v] > dfn[u]) ans[edge[x.id].id] = 2;
      } else
        cmin(low[u], dfn[x.v]);
    }
}
int Find(int x) { return fa[x] == x ? x : fa[x] = Find(fa[x]); }
int main() {
  int n, m;
  read(n, m);
  for (int i = 1; i <= (n); ++i) fa[i] = i;
  for (int i = 0; i < (m); ++i) {
    int u, v, w;
    read(u, v, w);
    edge.push_back(Edge(u, v, w, i));
  }
  sort(edge.begin(), edge.end());
  for (int i = 0; i < (m); ++i) {
    int nxt = i + 1;
    while (nxt != m && edge[nxt].w == edge[i].w) nxt++;
    for (int j = i; j < (nxt); ++j) {
      int x = Find(edge[j].u), y = Find(edge[j].v);
      if (x == y) continue;
      a[x].push_back(Node(y, j));
      a[y].push_back(Node(x, j));
      ans[edge[j].id] = 1;
    }
    for (int j = i; j < (nxt); ++j) {
      int x = Find(edge[j].u), y = Find(edge[j].v);
      if (x == y || dfn[x]) continue;
      Tarjan(x, -1);
    }
    for (int j = i; j < (nxt); ++j) {
      int x = Find(edge[j].u), y = Find(edge[j].v);
      if (x == y) continue;
      a[x].clear(), a[y].clear();
      dfn[x] = dfn[y] = 0;
      fa[x] = y;
    }
    i = nxt - 1;
  }
  for (int i = 0; i < (m); ++i)
    puts(ans[i] ? (ans[i] == 1 ? "at least one" : "any") : "none");
  return 0;
}
