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;
long long inf = 1000000000000000007;
long long mod = 1000000007, col = 1, n;
long long parent[100005];
vector<long long> adj[100005];
long long val[100005], ans[100005];
vector<pair<long long, long long>> v, edge;
long long dfs(long long node, long long par) {
  long long c = 0;
  parent[node] = par;
  for (auto u : adj[node]) {
    if (u != parent[node]) {
      c += dfs(u, node);
    }
  }
  return val[node] = c + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, x, y, m, z, i;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    edge.push_back({x, y});
  }
  z = dfs(1, 0);
  m = 0;
  for (auto u : edge) {
    x = u.first;
    y = u.second;
    if (y == parent[x]) {
      v.push_back({(val[x] * (n - val[x])), m});
    } else {
      v.push_back({(val[y] * (n - val[y])), m});
    }
    m++;
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n - 1; i++) {
    ans[v[i].second] = i;
  }
  for (i = 0; i < n - 1; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
