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 int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int INF = 1000000000;
const long long LINF = 1000000000000000000;
const long long MOD = 1000000007;
const double PI = acos(-1.0);
const double EPS = 1e-10;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<vector<int>> G(n);
  vector<int> ans(n, -1);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    G[u].push_back(i);
    G[v].push_back(i);
  }
  int cur = 0;
  for (int v = 0; v < n; v++) {
    if ((int)G[v].size() < 3) continue;
    for (auto e : G[v])
      if (ans[e] == -1) ans[e] = cur++;
  }
  for (int e = 0; e < n; e++)
    if (ans[e] == -1) ans[e] = cur++;
  for (int i = 0; i < n - 1; i++) cout << ans[i] << endl;
}
