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 = 100010, M = N << 1;
struct Edge {
  int a, b, w, type, id;
  bool operator<(const Edge &W) const { return w < W.w; }
} edges[N];
int h[N], e[M], num[M], ne[M], idx;
void add(int a, int b, int c) {
  e[idx] = b, num[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
int p[N];
int find(int x) {
  if (p[x] != x) p[x] = find(p[x]);
  return p[x];
}
int dfn[N], low[N], tot;
void tarjan(int u, int fa) {
  dfn[u] = low[u] = ++tot;
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i], k = num[i];
    if (k == fa) continue;
    if (!dfn[j]) {
      tarjan(j, k);
      low[u] = min(low[u], low[j]);
      if (low[j] > dfn[u]) edges[k].type = 2;
    } else
      low[u] = min(low[u], dfn[j]);
  }
}
int res[N];
int main() {
  memset(h, -1, sizeof h);
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) p[i] = i;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    edges[i] = (Edge){a, b, c, 0, i};
  }
  sort(edges, edges + m);
  for (int i = 0, j; i < m; i = j + 1) {
    j = i;
    while (j + 1 < m and edges[j + 1].w == edges[i].w) j++;
    for (int k = i; k <= j; k++) {
      int a = find(edges[k].a), b = find(edges[k].b);
      if (a != b) {
        add(a, b, k);
        add(b, a, k);
        edges[k].type = 1;
      }
    }
    for (int k = i; k <= j; k++) {
      int a = find(edges[k].a), b = find(edges[k].b);
      if (a != b and !dfn[a]) tarjan(a, -1);
    }
    for (int k = i; k <= j; k++) {
      int a = find(edges[k].a), b = find(edges[k].b);
      if (a != b) {
        h[a] = h[b] = -1;
        dfn[a] = dfn[b] = 0;
        p[a] = b;
      }
    }
  }
  for (int i = 0; i < m; i++) res[edges[i].id] = edges[i].type;
  for (int i = 0; i < m; i++) {
    if (!res[i])
      puts("none");
    else if (res[i] == 1)
      puts("at least one");
    else
      puts("any");
  }
  return 0;
}
