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;
inline long long read() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int maxn = 1e5 + 5;
struct node {
  int x, y;
} a[maxn];
int n, conn[maxn];
int main() {
  n = read();
  for (int i = 0; i < n - 1; ++i) {
    a[i].x = read();
    a[i].y = read();
    ++conn[a[i].x];
    ++conn[a[i].y];
  }
  int maxx = 0, id = 0;
  for (int i = 1; i <= n; ++i) {
    if (maxx < conn[i]) {
      maxx = conn[i];
      id = i;
    }
  }
  if (maxx == 1)
    printf("0\n");
  else if (maxx == 2) {
    int p = 0, q = 2;
    for (int i = 0; i < n - 1; ++i) {
      if (conn[a[i].x] == 1 || conn[a[i].y] == 1)
        printf("%d\n", p++);
      else
        printf("%d\n", q++);
    }
  } else {
    int p = 0, q = 3;
    for (int i = 0; i < n - 1; ++i) {
      if ((a[i].x == id || a[i].y == id) && p < 3)
        printf("%d\n", p++);
      else
        printf("%d\n", q++);
    }
  }
  return 0;
}
