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 lli = long long int;
using namespace std;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
class Prob_C {
 public:
  lli indegree = 0;
};
void solve() {
  lli n;
  cin >> n;
  Prob_C ind[n + 10];
  lli chk = 0, ele = 999999999, start = 0, finish = 3;
  vector<pair<lli, lli>> v;
  for (lli i = 0; i < n - 1; i++) {
    lli a;
    cin >> a;
    lli b;
    cin >> b;
    ind[a].indegree++;
    if (chk == 0 && ind[a].indegree >= 3) {
      chk = 1;
      ele = a;
    }
    ind[b].indegree++;
    if (chk == 0 && ind[b].indegree >= 3) {
      chk = 1;
      ele = b;
    }
    v.push_back(make_pair(a, b));
  }
  if (!chk) {
    for (lli i = 0; i < n - 1; i++) cout << i << endl;
  } else {
    for (lli i = 0; i < n - 1; i++) {
      lli a = v[i].first;
      lli b = v[i].second;
      if (a == ele || b == ele) {
        if (start < 3)
          cout << start++ << endl;
        else
          cout << finish++ << endl;
      } else
        cout << finish++ << endl;
    }
  }
}
int main() { solve(); }