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;
priority_queue<pair<long long, long long> > cancel;
long long siz[100005];
long long vis[100005];
vector<pair<long long, long long> > adj[100005];
long long weight[100005];
long long num[100005];
long long n;
void dfs(long long int src, long long int par, long long edgenum) {
  vis[src] = 1;
  for (long long i = 0; i < adj[src].size(); i++) {
    if (!vis[adj[src][i].first]) {
      dfs(adj[src][i].first, src, adj[src][i].second);
    }
  }
  siz[src]++;
  siz[par] += siz[src];
  if (edgenum != -1) weight[edgenum] = siz[src] * (n - siz[src]);
  return;
}
int main() {
  long long test = 1;
  for (long long z = 0; z < test; z++) {
    long long m, k;
    cin >> n;
    for (long long i = 0; i <= n; i++) {
      siz[i] = 0;
      vis[i] = 0;
    }
    for (long long i = 0; i < n - 1; i++) {
      cin >> m >> k;
      adj[m].push_back(make_pair(k, i));
      adj[k].push_back(make_pair(m, i));
    }
    dfs(1, 0, -1);
    for (long long i = 0; i < n - 1; i++) {
      cancel.push(make_pair(weight[i], i));
    }
    long long count = n - 2;
    while (!cancel.empty()) {
      num[cancel.top().second] = count;
      cancel.pop();
      count--;
    }
    for (long long i = 0; i < n - 1; i++) {
      cout << num[i] << endl;
    }
  }
}
