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 INF = 1LL << 60;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > e;
  vector<int> adj[n];
  for (int i = 0; i < n - 1; ++i) {
    int x1, x2;
    cin >> x1 >> x2;
    x1--;
    x2--;
    if (x1 > x2) swap(x1, x2);
    e.push_back({x1, x2});
    adj[x1].push_back(x2);
    adj[x2].push_back(x1);
  }
  bool bamboo = true;
  vector<pair<int, int> > special;
  for (int i = 0; i < n; ++i) {
    if ((int((adj[i]).size())) >= 3) {
      bamboo = false;
      for (int j = 0; j < 3; ++j) {
        int x = adj[i][j];
        if (x < i)
          special.push_back({x, i});
        else
          special.push_back({i, x});
      }
    }
  }
  if (bamboo) {
    for (int i = 0; i < n - 1; ++i) {
      cout << i << endl;
    }
  } else {
    int m = 3;
    for (int i = 0; i < n - 1; ++i) {
      bool said = false;
      for (int j = 0; j < 3; ++j) {
        if (e[i] == special[j]) {
          cout << j << endl;
          said = true;
        }
      }
      if (!said) {
        cout << m << endl;
        m++;
      }
    }
  }
}
int main() {
  solve();
  return 0;
}
