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;
struct tree {
  int lf;
  int rt;
  int we;
} a[2000010];
int v[2000010];
int main() {
  int n;
  scanf("%d", &n);
  for (int j = 0; j < n - 1; j++) {
    scanf("%d%d", &a[j].lf, &a[j].rt);
    a[j].we = -1;
    v[a[j].lf]++;
    v[a[j].rt]++;
  }
  int maxi = 0, ans = -1, p = 0;
  for (int j = 0; j < 2000010; j++) {
    if (maxi < v[j]) {
      maxi = v[j];
      ans = j;
    }
  }
  for (int j = 0; j < n - 1; j++)
    if ((a[j].lf == ans || a[j].rt == ans) && a[j].we == -1) a[j].we = p++;
  for (int j = 0; j < n - 1; j++)
    if (a[j].we == -1) a[j].we = p++;
  for (int j = 0; j < n - 1; j++) printf("%d\n", a[j].we);
  return 0;
}
