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;
vector<int> vec[200005];
struct node {
  int u, v, k;
  node(int uu, int vv, int kk) { u = uu, v = vv, k = kk; }
  bool operator<(const node& A) const {
    if (u == A.u) return v < A.v;
    return u < A.u;
  }
};
set<node> st;
int ans[200005];
int main() {
  int n;
  cin >> n;
  memset(ans, -1, sizeof(ans));
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
    st.insert(node(u, v, i));
  }
  bool wa = false;
  for (int i = 1; i <= n; i++) {
    if (vec[i].size() >= 3) {
      wa = true;
      for (int j = 0; j < 3; j++) {
        auto it = st.find(node(vec[i][j], i, 1));
        if (it != st.end()) {
          ans[it->k] = j;
        } else {
          it = st.find(node(i, vec[i][j], 1));
          ans[it->k] = j;
        }
      }
      break;
    }
  }
  int tmp = wa ? 3 : 0;
  for (int i = 1; i <= n - 1; i++) {
    if (ans[i] == -1) {
      ans[i] = tmp++;
    }
  }
  for (int i = 1; i <= n - 1; i++) cout << ans[i] << endl;
  return 0;
}
