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>
inline void rd(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') f = ch == '-' ? -1 : 1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  x *= f;
}
template <class T>
inline void ckmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void ckmin(T &a, T b) {
  if (a > b) a = b;
}
const int N = 100005;
int n, x, y, ans[N], d[N];
vector<int> g[N];
bool vis[N];
int main() {
  rd(n);
  if (n == 2) puts("0"), exit(0);
  if (n == 3) puts("0"), puts("1"), exit(0);
  for (int i = 1, I = (n - 1) + 1; i < I; ++i)
    rd(x), rd(y), g[x].push_back(i), g[y].push_back(i), ++d[x], ++d[y];
  int u = max_element(d + 1, d + n + 1) - d;
  int cnt = 0;
  ans[g[u][0]] = ++cnt, ans[g[u][1]] = ++cnt;
  if (g[u].size() > 2) ans[g[u][2]] = ++cnt;
  for (int i = 1, I = (n - 1) + 1; i < I; ++i)
    if (!ans[i]) ans[i] = ++cnt;
  for (int i = 1, I = (n - 1) + 1; i < I; ++i) printf("%d\n", ans[i] - 1);
}