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;
struct node {
  int x, y, next, h;
} edge[2 * 210000];
int first[210000], len;
struct Edge {
  int x, y, h, d;
  bool operator<(const Edge &B) const { return d < B.d; }
} g[210000];
int n, m;
int ans[210000], father[210000], dfn[210000], low[210000], cnt;
bool mark[210000];
vector<int> Set;
void ins(int x, int y, int h) {
  edge[++len] = (node){x, y, first[x], h};
  first[x] = len;
}
void init() {
  int i, j, k;
  memset(first, -1, sizeof(first));
  len = 0;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d", &g[i].x, &g[i].y, &g[i].d);
    g[i].h = i;
  }
}
int get_fa(int x) {
  if (father[x] == x) return x;
  return father[x] = get_fa(father[x]);
}
void tarjan(int x) {
  int i, j, k, y;
  dfn[x] = low[x] = ++cnt;
  for (k = first[x]; k != -1; k = edge[k].next) {
    y = edge[k].y;
    if (mark[k / 2]) continue;
    mark[k / 2] = 1;
    if (!dfn[y]) {
      tarjan(y);
      if (dfn[x] < low[y]) {
        ans[edge[k].h] = 1;
      }
      low[x] = (low[x] < low[y] ? low[x] : low[y]);
    } else {
      low[x] = (dfn[y] < low[x] ? dfn[y] : low[x]);
    }
  }
}
void solve() {
  int i, j, k, head, tail;
  memset(ans, 0, sizeof(ans));
  memset(dfn, 0, sizeof(dfn));
  cnt = 0;
  memset(first, -1, sizeof(first));
  len = -1;
  memset(mark, 0, sizeof(mark));
  for (i = 1; i <= n; i++) father[i] = i;
  sort(g + 1, g + 1 + m);
  for (head = 1; head <= m; head++) {
    Set.clear();
    for (tail = head + 1; tail <= m; tail++) {
      if (g[tail].d != g[head].d) break;
    }
    tail--;
    for (i = head; i <= tail; i++) {
      int x = get_fa(g[i].x), y = get_fa(g[i].y);
      if (x == y) {
        ans[g[i].h] = -1;
        continue;
      }
      Set.push_back(x);
      ins(x, y, g[i].h);
      ins(y, x, g[i].h);
    }
    while (!Set.empty()) {
      i = Set.back();
      Set.pop_back();
      if (!dfn[i]) tarjan(i);
    }
    k = 0;
    for (i = head; i <= tail; i++) {
      if (ans[g[i].h] == -1) continue;
      int x = get_fa(g[i].x), y = get_fa(g[i].y);
      father[x] = y;
      first[x] = -1;
      first[y] = -1;
      mark[k] = 0;
      dfn[x] = dfn[y] = 0;
      k++;
    }
    head = tail;
    len = -1;
    cnt = 0;
  }
  for (i = 1; i <= m; i++) {
    if (ans[i] == -1) puts("none");
    if (ans[i] == 0) puts("at least one");
    if (ans[i] == 1) puts("any");
  }
}
int main() {
  init();
  solve();
  return 0;
}
