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 maxn = 1e5 + 5;
int n, m, cnt = 0, dfn[maxn], low[maxn], father[maxn];
vector<pair<int, int>> g[maxn];
struct node {
  int u, v, w, id, ans;
} s[maxn];
bool cmp(const node &a, const node &b) { return a.w < b.w; }
bool cmp1(const node &a, const node &b) { return a.id < b.id; }
int findfather(int x) {
  if (x == father[x]) return x;
  int i = findfather(father[x]);
  father[x] = i;
  return i;
}
void tarjan(int x, int fa) {
  low[x] = dfn[x] = ++cnt;
  for (auto to : g[x]) {
    if (fa == to.second) continue;
    if (!dfn[to.first]) {
      tarjan(to.first, to.second);
      low[x] = min(low[x], low[to.first]);
      if (low[to.first] > dfn[x]) s[to.second].ans = 2;
    } else
      low[x] = min(low[x], low[to.first]);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; ++i) father[i] = i;
  for (int i = 1; i <= m; ++i)
    scanf("%d%d%d", &s[i].u, &s[i].v, &s[i].w), s[i].id = i;
  sort(s + 1, s + 1 + m, cmp);
  int i = 1, j;
  while (i <= m) {
    j = i;
    while (j <= m && s[i].w == s[j].w) j++;
    for (int k = i; k < j; ++k) {
      int fa = findfather(s[k].u), fb = findfather(s[k].v);
      if (fa == fb) continue;
      g[fa].push_back({fb, k});
      g[fb].push_back({fa, k});
      s[k].ans = 1;
    }
    for (int k = i; k < j; ++k) {
      int fa = findfather(s[k].u), fb = findfather(s[k].v);
      if (fa == fb || dfn[fa]) continue;
      tarjan(fa, -1);
    }
    for (int k = i; k < j; ++k) {
      int fa = findfather(s[k].u), fb = findfather(s[k].v);
      if (fa == fb) continue;
      g[fa].clear();
      g[fb].clear();
      dfn[fa] = dfn[fb] = 0;
      father[fa] = fb;
    }
    i = j;
  }
  sort(s + 1, s + 1 + m, cmp1);
  for (int i = 1; i <= m; ++i)
    if (s[i].ans == 0)
      printf("none\n");
    else if (s[i].ans == 1)
      printf("at least one\n");
    else
      printf("any\n");
}
