You are given a tree consisting of n nodes. You want to write some labels on the tree's edges such that the following conditions hold:

  * Every label is an integer between 0 and n-2 inclusive. 
  * All the written labels are distinct. 
  * The largest value among MEX(u,v) over all pairs of nodes (u,v) is as small as possible. 



Here, MEX(u,v) denotes the smallest non-negative integer that isn't written on any edge on the unique simple path from node u to node v.

Input

The first line contains the integer n (2 ≤ n ≤ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines contains two space-separated integers u and v (1 ≤ u,v ≤ n) that mean there's an edge between nodes u and v. It's guaranteed that the given graph is a tree.

Output

Output n-1 integers. The i^{th} of them will be the number written on the i^{th} edge (in the input order).

Examples

Input


3
1 2
1 3


Output


0
1


Input


6
1 2
1 3
2 4
2 5
5 6


Output


0
3
2
4
1

Note

The tree from the second sample:

<image>
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 100000;
struct Edge {
  long long u, i;
};
vector<Edge> graph[MAXN];
long long colors[MAXN];
long long dist[MAXN];
long long weig[MAXN];
long long indx[MAXN];
vector<long long> lifts;
void calcd(long long v, long long p, long long d) {
  bool islift = true;
  dist[v] = d;
  weig[v] = 1;
  for (auto u : graph[v]) {
    if (u.u == p) continue;
    islift = false;
    indx[u.u] = u.i;
    calcd(u.u, v, d + 1);
    weig[v] += weig[u.u];
  }
  if (islift) lifts.push_back(v);
}
bool cmp(const long long a, const long long b) { return weig[a] < weig[b]; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n == 2) {
    cout << 0;
    return 0;
  }
  if (n == 3) {
    cout << "0\n1";
    return 0;
  }
  fill(colors, colors + n - 1, -1);
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    graph[a].push_back({b, i});
    graph[b].push_back({a, i});
  }
  long long root = 0;
  while (graph[root].size() <= 1) root++;
  calcd(root, -1, 0);
  if (lifts.size() <= 2) {
    for (long long i = 0; i < n - 1; ++i) cout << i << ' ';
    return 0;
  }
  for (long long i = 0; i < lifts.size(); ++i) colors[indx[lifts[i]]] = i;
  long long clr = lifts.size();
  for (long long i = 0; i < n - 1; ++i) {
    if (colors[i] == -1) colors[i] = clr++;
  }
  for (long long i = 0; i < n - 1; ++i) cout << colors[i] << '\n';
  return 0;
}
