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 namespace std;
vector<long long> answer;
struct TreeNode {
  long long nvalue;
  struct TreeNode* left;
  struct TreeNode* right;
};
long long globalCount = 0;
void generateTrees(string s, long long direction, long long E, long long total,
                   vector<string>& alls) {
  if (total <= 0) return;
  long long i;
  if (direction == 0) {
    s += 'L';
  } else {
    s += 'R';
  }
  if (E == total - 1) {
    alls.push_back(s);
    return;
  }
  for (i = 0; i <= 1; i++) {
    generateTrees(s, i, E + 1, total, alls);
  }
}
long long ar[2 * 1000050] = {0};
void prime() {
  long long i, j;
  for (i = 2; i < 2 * 1000050; i++) {
    if (ar[i] == 1) continue;
    for (j = 2 * i; j < 2 * 1000050; j += i) {
      ar[j] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long i;
  map<pair<long long, long long>, long long> answer;
  vector<long long> e[n + 1];
  vector<pair<long long, long long> > givenOrder;
  for (i = 0; i <= n - 2; i++) {
    long long x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
    givenOrder.push_back({x, y});
  }
  long long count = 0;
  for (i = 0; i < n - 1; i++) {
    pair<long long, long long> edge = givenOrder[i];
    if (e[edge.first].size() == 1 || e[edge.second].size() == 1) {
      answer[edge] = count;
      count++;
    } else {
      answer[edge] = -1;
    }
  }
  for (i = 0; i < n - 1; i++) {
    pair<long long, long long> edge = givenOrder[i];
    if (answer[edge] == -1) {
      answer[edge] = count;
      count++;
    }
  }
  for (i = 0; i < n - 1; i++) {
    cout << answer[givenOrder[i]] << "\n";
  }
  return 0;
}