Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> G[3000000 + 5];
pair<int, long long> table[3000000 + 5];
int n, size[3000000 + 5] = {}, ans[3000000 + 5] = {}, fa[3000000 + 5] = {};
int input[3000000 + 5][2];
bool cmp(pair<int, long long> x, pair<int, long long> y) {
  return x.second < y.second;
}
void dfs(int Fa, int x) {
  fa[x] = Fa;
  for (int i = 0; i < G[x].size(); ++i) {
    int v = G[x][i];
    if (v != Fa) {
      dfs(x, v);
      size[x] += size[v];
    }
  }
  ++size[x];
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) G[i].clear();
  for (int i = 1; i < n; ++i) {
    scanf("%d %d", &input[i][0], &input[i][1]);
    G[input[i][1]].push_back(input[i][0]);
    G[input[i][0]].push_back(input[i][1]);
  }
  dfs(0, 1);
  for (int i = 1; i < n; ++i) {
    int x = input[i][0], y = input[i][1];
    table[i].first = i;
    if (fa[x] == y)
      table[i].second = (long long)size[x] * (n - size[x]);
    else {
      table[i].second = (long long)size[y] * (n - size[y]);
    }
  }
  sort(table + 1, table + n, cmp);
  for (int i = 1; i < n; ++i) ans[table[i].first] = i;
  for (int i = 1; i < n; ++i) printf("%d\n", ans[i] - 1);
  return 0;
}