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 = 3e5 + 10;
struct ds {
  int u, v, c, res, vt;
};
ds a[N];
bool cmp(ds a, ds b) { return a.c < b.c; }
int n, m, Count = 0, number[N], low[N], p[N], dd[N], kt[N];
vector<int> stk[N], b[N];
void nhap() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].c);
    a[i].vt = i;
  }
  sort(a + 1, a + 1 + m, cmp);
}
void init() {
  for (int i = 1; i <= n; i++) {
    p[i] = -1;
  }
}
int find_root(int u) {
  if (p[u] == -1) {
    return u;
  } else {
    p[u] = find_root(p[u]);
    return p[u];
  }
}
void dfs(int u) {
  Count++;
  low[u] = Count;
  number[u] = Count;
  dd[u] = 1;
  for (int i = 0; i < stk[u].size(); i++) {
    int v = stk[u][i];
    int vt = b[u][i];
    if (kt[vt] == 0) {
      kt[vt] = 1;
      if (dd[v] == 0) {
        dfs(v);
        if (low[v] > number[u]) {
          a[vt].res = 2;
        }
        low[u] = min(low[u], low[v]);
      } else {
        low[u] = min(low[u], number[v]);
      }
    }
  }
}
void them_canh(int u, int v, int i) {
  int x = find_root(u);
  int y = find_root(v);
  stk[x].push_back(y);
  stk[y].push_back(x);
  b[x].push_back(i);
  b[y].push_back(i);
  if (x == y) {
    return;
  }
  a[i].res = 1;
}
void reset(int u, int v) {
  int x = find_root(u);
  int y = find_root(v);
  dd[x] = dd[y] = 0;
  stk[x].clear();
  b[x].clear();
  stk[y].clear();
  b[y].clear();
}
void ghep(int u, int v) {
  int x = find_root(u);
  int y = find_root(v);
  if (x == y) {
    return;
  }
  p[x] = y;
}
bool cmp1(ds a, ds b) { return a.vt < b.vt; }
void solve() {
  int i = 1;
  while (i <= m) {
    int j;
    Count = 0;
    for (j = i; a[i].c == a[j].c; j++) {
      them_canh(a[j].u, a[j].v, j);
    }
    for (j = i; a[i].c == a[j].c; j++) {
      int u = find_root(a[j].u);
      int v = find_root(a[j].v);
      if (dd[u] == 0) {
        dfs(u);
      }
      if (dd[v] == 0) {
        dfs(v);
      }
    }
    for (j = i; a[i].c == a[j].c; j++) {
      reset(a[j].u, a[j].v);
    }
    for (j = i; a[i].c == a[j].c; j++) {
      ghep(a[j].u, a[j].v);
    }
    i = j;
  }
  sort(a + 1, a + 1 + m, cmp1);
  for (int i = 1; i <= m; i++) {
    if (a[i].res == 0) {
      printf("none\n");
    } else {
      printf("%s\n", a[i].res == 1 ? "at least one" : "any");
    }
  }
}
int main() {
  nhap();
  init();
  solve();
}
