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 INF = 1e9;
const int MOD = 1e9 + 7;
struct Edge {
  int x, y, no;
};
vector<Edge> arr;
int deg[101010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    arr.push_back({x, y, -1});
    deg[x]++;
    deg[y]++;
  }
  int d3 = -1;
  for (int i = 1; i <= n; i++) {
    if (deg[i] >= 3) d3 = i;
  }
  if (d3 == -1) {
    for (int i = 0; i < n - 1; i++) printf("%d\n", i);
  } else {
    int cnt = 0;
    for (int i = 0; i < arr.size(); i++) {
      if (arr[i].x == d3 || arr[i].y == d3) arr[i].no = cnt++;
      if (cnt >= 3) break;
    }
    for (int i = 0; i < n - 1; i++) {
      if (arr[i].no == -1) arr[i].no = cnt++;
      printf("%d\n", arr[i].no);
    }
  }
  return 0;
}
