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 = 201010;
const int E = 201010;
const char meg[3][15] = {"none", "any", "at least one"};
struct Edge {
  int x, y, c;
  int id;
  int nxt;
} bf[E], rec[E];
int ne;
int head[N];
void addedge(int x, int y, int c, int id) {
  bf[ne].x = x;
  bf[ne].y = y;
  bf[ne].c = c;
  bf[ne].id = id;
  bf[ne].nxt = head[x];
  head[x] = ne++;
}
int p[N];
int n, m;
int ans[E];
void init() {
  int i;
  for (i = 1; i <= n; i++) p[i] = i;
}
int findp(int x) {
  if (p[x] != x) p[x] = findp(p[x]);
  return p[x];
}
int dfn[N];
int low[N];
int dindex;
void tarjan(int u, int use) {
  dfn[u] = low[u] = dindex++;
  int i;
  for (i = head[u]; i; i = bf[i].nxt) {
    if (bf[i].id == use) continue;
    int v = bf[i].y;
    if (dfn[v]) {
      low[u] = min(low[u], dfn[v]);
    } else {
      tarjan(v, bf[i].id);
      low[u] = min(low[u], low[v]);
      if (low[v] > dfn[u]) ans[bf[i].id] = 1;
    }
  }
}
bool cmp(const Edge &A, const Edge &B) { return A.c < B.c; }
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    int i, j;
    for (i = 0; i < m; i++) {
      int x, y, c;
      scanf("%d%d%d", &x, &y, &c);
      rec[i].x = x;
      rec[i].y = y;
      rec[i].c = c;
      rec[i].id = i;
    }
    rec[m].c = 1000000000;
    sort(rec, rec + m, cmp);
    init();
    vector<int> u, rr;
    for (i = 0; i < m; i++) {
      u.push_back(i);
      if (rec[i + 1].c == rec[u[0]].c) continue;
      ne = 2;
      for (j = 0; j < u.size(); j++) {
        int xx = findp(rec[u[j]].x);
        int yy = findp(rec[u[j]].y);
        if (xx == yy) {
          ans[rec[u[j]].id] = 0;
          continue;
        }
        ans[rec[u[j]].id] = 2;
        rr.push_back(xx);
        rr.push_back(yy);
        addedge(xx, yy, rec[u[j]].c, rec[u[j]].id);
        addedge(yy, xx, rec[u[j]].c, rec[u[j]].id);
      }
      dindex = 0;
      for (j = 0; j < rr.size(); j++)
        if (!dfn[rr[j]]) {
          tarjan(rr[j], -1);
        }
      for (j = 0; j < u.size(); j++) {
        int xx = findp(rec[u[j]].x);
        int yy = findp(rec[u[j]].y);
        if (xx != yy) p[xx] = yy;
      }
      for (j = 0; j < rr.size(); j++) {
        head[rr[j]] = 0;
        dfn[rr[j]] = low[rr[j]] = 0;
      }
      rr.clear();
      u.clear();
    }
    for (i = 0; i < m; i++) puts(meg[ans[i]]);
  }
}
