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 N = 100005;
struct ed {
  int x, y, z, id;
  bool operator<(const ed &o) const { return z < o.z; }
} a[N];
int n, m;
int tot, lnk[N], nxt[N << 1], son[N << 1], id[N << 1];
int clo, fa[N], dfn[N], low[N], ans[N];
void add(int x, int y, int i) {
  nxt[++tot] = lnk[x], lnk[x] = tot, son[tot] = y, id[tot] = i;
}
void tarjan(int x, int p) {
  low[x] = dfn[x] = ++clo;
  for (int j = lnk[x], y; j; j = nxt[j]) {
    if (!dfn[y = son[j]]) {
      tarjan(y, id[j]), low[x] = min(low[x], low[y]);
      if (low[y] > dfn[x]) ans[id[j]] = 1;
    } else if (dfn[x] > dfn[y] && id[j] != p)
      low[x] = min(low[x], dfn[y]);
  }
}
int get(int x) { return fa[x] == x ? x : fa[x] = get(fa[x]); }
void unn(int x, int y) {
  int fx = get(x), fy = get(y);
  if (fx != fy) {
    fa[fx] = fy, tot = 0;
    dfn[fx] = dfn[fy] = 0;
    lnk[fx] = lnk[fy] = 0;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) fa[i] = i;
  for (int i = 1; i <= m; ++i)
    scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z), a[i].id = i;
  sort(a + 1, a + 1 + m);
  for (int i = 1, j; i <= m; ++i) {
    for (j = i + 1; j <= m && a[i].z == a[j].z; ++j)
      ;
    for (int k = i, fx, fy; k < j; ++k) {
      fx = get(a[k].x), fy = get(a[k].y);
      if (fx != fy) {
        add(fx, fy, a[k].id), add(fy, fx, a[k].id);
        ans[a[k].id] = 2;
      }
    }
    for (int k = i, fx, fy; k < j; ++k) {
      fx = get(a[k].x), fy = get(a[k].y);
      if (fx != fy && !dfn[fx]) clo = 0, tarjan(fx, 0);
    }
    for (int k = i; k < j; ++k) unn(a[k].x, a[k].y);
    i = j - 1;
  }
  for (int i = 1; i <= m; ++i) {
    if (ans[i] == 0)
      puts("none");
    else if (ans[i] == 1)
      puts("any");
    else
      puts("at least one");
  }
  return 0;
}
