Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long int mod = 1000000007;
int n;
vector<vector<long long> > edges;
vector<pair<long long, long long> > edgeList;
map<pair<long long, long long>, long long> mp;
vector<pair<long long, long long> > deg;
int num = 0;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  deg.resize(n);
  edges.resize(n);
  edgeList.resize(n);
  for (long long i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    deg[u].second++;
    deg[v].second++;
    deg[u].first = u;
    deg[v].first = v;
    edges[u].push_back(v);
    edges[v].push_back(u);
    edgeList[i] = {u, v};
  }
  sort(deg.begin(), deg.end(),
       [&](pair<long long, long long> a, pair<long long, long long> b) {
         return a.second > b.second;
       });
  int pos = deg[0].first;
  queue<int> q;
  q.push(pos);
  vector<long long> visited(n, 0);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    visited[u] = 1;
    for (auto it : edges[u]) {
      if (visited[it] == 0) {
        mp[{u, it}] = num++;
        q.push(it);
      }
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (mp.find(edgeList[i]) != mp.end()) {
      cout << mp[edgeList[i]] << '\n';
    } else {
      cout << mp[{edgeList[i].second, edgeList[i].first}] << '\n';
    }
  }
  return 0;
}