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;
int t, N, x, a, b, n;
int A[100000];
int ans[100000];
vector<pair<long long, long long> > V[100000];
int ck[100000];
int cnt = 0;
unordered_set<int> S;
void dfs(int a) {
  ck[a] = 1;
  for (pair<long long, long long> p : V[a]) {
    int i = p.first;
    if (ck[i]) continue;
    ans[p.second] = cnt;
    cnt++;
    dfs(i);
    return;
  }
}
int main() {
  cin >> n;
  for (int i = 0, a, b; i < n - 1; i++) {
    cin >> a >> b;
    V[a - 1].push_back(pair<long long, long long>(b - 1, i));
    V[b - 1].push_back(pair<long long, long long>(a - 1, i));
  }
  int a = -1;
  for (int i = 0; i < n; i++) {
    if (V[i].size() == 1 && a == -1) {
      a = i;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    ans[i] = -1;
  }
  dfs(a);
  if (cnt == n - 1) {
    for (int i = 0; i < n - 1; i++) {
      cout << i << "\n";
    }
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    if (ans[i] == -1) {
      continue;
    }
    if (ans[i] == cnt - 1) {
      ans[i] = 1;
    } else if (ans[i] == 0) {
      continue;
    } else {
      ans[i] += 2;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    S.insert(i);
  }
  for (int i = 0; i < n - 1; i++) {
    if (ans[i] >= 0) {
      S.erase(ans[i]);
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (ans[i] == -1) {
      int a = *S.begin();
      ans[i] = a;
      S.erase(a);
    }
  }
  for (int i = 0; i < n - 1; i++) {
    cout << ans[i] << "\n";
  }
}