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;
template <class T>
using reversed_priority_queue = priority_queue<T, vector<T>, greater<T> >;
int n;
vector<int> G[100010];
int degree[100010];
bool used[100010];
map<pair<int, int>, int> idx;
map<pair<int, int>, int> used_e;
int vv[100010];
int uv[100010];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[v].push_back(u);
    G[u].push_back(v);
    degree[v]++;
    degree[u]++;
    vv[i] = v;
    uv[i] = u;
    idx[make_pair(min(u, v), max(u, v))] = i;
  }
  queue<int> que;
  for (int i = 1; i <= n; i++) {
    if (degree[i] == 1) {
      que.push(i);
    }
  }
  int cnt = 0;
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    if (used[v]) continue;
    used[v] = true;
    for (int i = 0; i < G[v].size(); i++) {
      int to = G[v][i];
      pair<int, int> pt = make_pair(min(v, to), max(v, to));
      if (used_e.find(pt) == used_e.end()) {
        used_e[pt] = cnt;
        cnt++;
      }
      que.push(to);
    }
  }
  for (int i = 0; i < n - 1; i++) {
    pair<int, int> pt = make_pair(min(uv[i], vv[i]), max(uv[i], vv[i]));
    cout << used_e[pt] << endl;
  }
  return 0;
}