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 = 100005;
vector<long long> g[maxn];
vector<long long> deg(maxn);
set<long long> mex;
map<pair<long long, long long>, long long> ans, done;
vector<pair<long long, long long> > edges;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i <= n - 2; i++) {
    mex.insert(i);
  }
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    deg[u]++;
    deg[v]++;
    edges.push_back({u, v});
  }
  vector<pair<long long, long long> > pairs;
  for (long long i = 1; i <= n; i++) {
    pairs.push_back({deg[i], i});
  }
  sort(pairs.begin(), pairs.end());
  for (long long i = 0; i < n; i++) {
    long long cur = pairs[i].second;
    for (long long nxt : g[cur]) {
      if (done[{nxt, cur}] || done[{cur, nxt}]) continue;
      ans[{nxt, cur}] = ans[{cur, nxt}] = *mex.begin();
      mex.erase(mex.begin());
      done[{nxt, cur}] = done[{cur, nxt}] = 1;
    }
  }
  for (pair<long long, long long> it : edges) {
    cout << ans[it] << '\n';
  }
}
signed main() {
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
