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 = 1e5 + 10;
vector<pair<long long, long long>> g[maxn];
pair<long long, long long> pr[maxn];
vector<long long> lists;
void dfs(long long v, long long p, long long ed) {
  if (g[v].size() == 1) {
    lists.push_back(ed);
  }
  for (auto to : g[v]) {
    if (to.first == p) continue;
    dfs(to.first, v, to.second);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  long long st[n];
  for (long long j = 0; j < n; ++j) {
    st[j] = 0;
  }
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back({b, i});
    g[b].push_back({a, i});
    st[a]++;
    st[b]++;
    pr[a] = {b, i};
    pr[b] = {a, i};
  }
  long long cnt = 0;
  for (long long k = 0; k < n; ++k) {
    if (st[k] == 1) {
      cnt++;
    }
  }
  long long list = 0;
  if (cnt == 2) {
    for (long long i = 0; i < n - 1; ++i) {
      cout << i << "\n";
    }
    exit(0);
  }
  for (long long l = 0; l < n; ++l) {
    if (st[l] == 1) list = l;
  }
  long long root = pr[list].first;
  dfs(root, list, -1);
  long long ans[n - 1];
  for (long long m = 0; m < n - 1; ++m) {
    ans[m] = -1;
  }
  ans[pr[list].second] = 0;
  ans[lists[0]] = 1;
  ans[lists[1]] = 2;
  cnt = 3;
  for (long long i1 = 0; i1 < n - 1; ++i1) {
    if (ans[i1] == -1) {
      ans[i1] = cnt;
      cnt++;
    }
  }
  for (long long i = 0; i < n - 1; ++i) {
    cout << ans[i] << "\n";
  }
}
