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 <typename T>
void read(T& x) {
  x = 0;
  int fl = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fl = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  x *= fl;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
const int N = 100005;
int n, ans[N], now;
bool b[N];
struct Edge {
  int nxt, to, id;
} edge[N << 1];
int head[N], edgeNum, deg[N];
void addEdge(int from, int to, int id) {
  edge[++edgeNum] = (Edge){head[from], to, id};
  head[from] = edgeNum;
  deg[to]++;
}
int main() {
  memset(ans, -1, sizeof ans);
  read(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    read(u, v);
    addEdge(u, v, i), addEdge(v, u, i);
  }
  int mark = 0;
  for (int i = 1; i <= n; i++)
    if (deg[i] > 2) mark = i;
  if (mark)
    for (int i = head[mark]; i; i = edge[i].nxt) ans[edge[i].id] = now++;
  for (int i = 1; i < n; i++) {
    if (ans[i] == -1) ans[i] = now++;
    printf("%d\n", ans[i]);
  }
  return 0;
}