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;
const int maxn = 1e5 + 5;
int n;
int U[maxn], V[maxn];
vector<int> adj[maxn];
map<pair<int, int>, int> used;
vector<int> P[maxn];
void dfs(int u, int par) {
  for (int v : adj[u]) {
    if (v == par) continue;
    P[u].push_back(v);
    dfs(v, u);
  }
}
void gan(int u, int v, int val) {
  int x = min(u, v);
  int y = max(u, v);
  used[make_pair(x, y)] = val;
}
void solve() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> U[i] >> V[i];
    adj[U[i]].push_back(V[i]);
    adj[V[i]].push_back(U[i]);
  }
  dfs(1, 0);
  bool found = false;
  for (int u = 1; u <= n; u++) {
    if (adj[u].size() >= 3) {
      int ver = -1;
      for (int v : adj[u]) {
        if (adj[v].size() > 1) {
          found = true;
          ver = v;
          break;
        }
      }
      if (ver != -1) {
        int dem = 0;
        for (int v : adj[u]) {
          if (v != ver && dem < 2) {
            dem++;
            if (dem == 1)
              gan(u, v, 1);
            else
              gan(u, v, 3);
          }
        }
        for (int cc : adj[ver]) {
          if (cc != u) {
            gan(ver, cc, 2);
            break;
          }
        }
      }
    }
    if (found) break;
  }
  int cur;
  if (found)
    cur = 3;
  else
    cur = 0;
  for (int i = 1; i < n; i++) {
    int x = min(U[i], V[i]);
    int y = max(U[i], V[i]);
    if (!used[make_pair(x, y)]) {
      cur++;
      gan(x, y, cur);
    }
  }
  for (int i = 1; i < n; i++) {
    int x = min(U[i], V[i]);
    int y = max(U[i], V[i]);
    cout << used[make_pair(x, y)] - 1 << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}