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;
vector<int> G[100005], id[100005];
int d[100005], ans[100005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    d[u]++;
    d[v]++;
    G[u].push_back(v);
    G[v].push_back(u);
    id[u].push_back(i);
    id[v].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] >= 3) {
      ans[id[i][0]] = 1;
      ans[id[i][1]] = 2;
      ans[id[i][2]] = 3;
      int t = 3;
      for (int j = 1; j < n; j++)
        if (!ans[j]) ans[j] = ++t;
      for (int j = 1; j < n; j++) printf("%d\n", ans[j] - 1);
      return 0;
    }
  }
  for (int i = 1; i < n; i++) printf("%d\n", i - 1);
  return 0;
}
