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 Z>
inline bool checkmax(Z &a, Z b) {
  if (a == -1 || a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class Z>
inline bool checkmin(Z &a, Z b) {
  if (a == -1 || a > b) {
    a = b;
    return true;
  }
  return false;
}
int n, m;
struct Road {
  int u, v, len;
  int idx;
  bool operator<(const Road &b) const { return len < b.len; }
} road[1100000];
int tree[110000];
int find(int x) {
  if (x == tree[x]) return x;
  return tree[x] = find(tree[x]);
}
int ans[110000];
int Dindex, Bcnt, Stop, idx;
int LOW[110000];
int DFN[110000];
int Belong[110000];
int stap[110000];
int instack[110000];
int isbridge[110000];
int net[110000], size;
struct EDGE {
  int v, next;
  int idx;
} edge[1100000];
void init() {
  size = 0;
  memset(net, -1, sizeof(net));
}
void add(int u, int v, int idx) {
  edge[size].v = v;
  edge[size].idx = idx;
  edge[size].next = net[u];
  net[u] = size++;
}
void tarjan(int u, int fa) {
  DFN[u] = LOW[u] = ++Dindex;
  stap[++Stop] = u;
  int ff = 1;
  for (int i = net[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (v == fa && ff) {
      ff = 0;
      continue;
    }
    if (!DFN[v]) {
      tarjan(v, u);
      if (LOW[v] < LOW[u]) LOW[u] = LOW[v];
      if (LOW[v] > DFN[u]) {
        ans[edge[i].idx] = 2;
      }
    } else if (DFN[v] < LOW[u]) {
      LOW[u] = DFN[v];
    }
  }
}
void solve() {
  Dindex = Bcnt = Stop = 0;
  memset(DFN, 0, sizeof(DFN));
  for (int i = 1; i <= n; i++) {
    if (!DFN[i]) tarjan(i, -1);
  }
}
int main() {
  int i, j;
  while (scanf("%d%d", &n, &m) != EOF) {
    init();
    for (i = 0; i <= n; i++) tree[i] = i;
    for (i = 1; i <= m; i++) {
      scanf("%d%d%d", &road[i].u, &road[i].v, &road[i].len);
      road[i].idx = i;
      ans[i] = 0;
    }
    sort(road + 1, road + 1 + m);
    for (i = 1; i <= m;) {
      for (j = i; j <= m && road[i].len == road[j].len; j++) {
        int fa = find(road[j].u);
        int fb = find(road[j].v);
        if (fa != fb) {
          ans[road[j].idx]++;
          add(fa, fb, road[j].idx);
          add(fb, fa, road[j].idx);
        }
      }
      int k = 0;
      for (j = i; j <= m && road[i].len == road[j].len; j++) {
        int fa = find(road[j].u);
        int fb = find(road[j].v);
        if (fa != fb) {
          k++;
          tree[fa] = fb;
        }
      }
      i = j;
    }
    solve();
    for (i = 1; i <= m; i++) {
      if (ans[i] == 0)
        puts("none");
      else if (ans[i] == 1)
        puts("at least one");
      else
        puts("any");
    }
  }
  return 0;
}
