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 MAXN = 114514;
int N, M;
int T;
void YES() { printf("YES\n"); }
void NO() { printf("NO\n"); }
void read(int &a) { scanf("%d", &a); }
int out[MAXN];
int p[MAXN];
int answer[MAXN];
void work() {
  scanf("%d", &N);
  int f, t;
  for (int i = 1; i <= N - 1; ++i) {
    scanf("%d%d", &f, &t);
    out[f]++;
    out[t]++;
    p[f] = p[t] = i;
  }
  if (N == 2) {
    printf("%d", 0);
    return;
  }
  if (N == 3) {
    printf("%d\n%d", 0, 1);
    return;
  }
  int sum = 0;
  for (int i = 1; i <= N; ++i) {
    if (out[i] == 1) {
      sum++;
    }
  }
  if (sum == 2) {
    for (int i = 0; i <= N - 2; ++i) {
      printf("%d\n", i);
    }
    return;
  }
  int cnt = 0;
  for (int i = 1; i <= N; ++i) {
    if (out[i] == 1) {
      answer[p[i]] = ++cnt;
      if (cnt == 3) {
        break;
      }
    }
  }
  for (int i = 1; i <= N - 1; ++i) {
    if (answer[i]) {
      printf("%d\n", answer[i] - 1);
    } else {
      ++cnt;
      printf("%d\n", cnt - 1);
    }
  }
}
int main() {
  work();
  return 0;
}
