Evlampiy was gifted a rooted tree. The vertices of the tree are numbered from 1 to n. Each of its vertices also has an integer a_i written on it. For each vertex i, Evlampiy calculated c_i — the number of vertices j in the subtree of vertex i, such that a_j < a_i. 

<image>Illustration for the second example, the first integer is a_i and the integer in parentheses is c_i

After the new year, Evlampiy could not remember what his gift was! He remembers the tree and the values of c_i, but he completely forgot which integers a_i were written on the vertices.

Help him to restore initial integers!

Input

The first line contains an integer n (1 ≤ n ≤ 2000) — the number of vertices in the tree.

The next n lines contain descriptions of vertices: the i-th line contains two integers p_i and c_i (0 ≤ p_i ≤ n; 0 ≤ c_i ≤ n-1), where p_i is the parent of vertex i or 0 if vertex i is root, and c_i is the number of vertices j in the subtree of vertex i, such that a_j < a_i.

It is guaranteed that the values of p_i describe a rooted tree with n vertices.

Output

If a solution exists, in the first line print "YES", and in the second line output n integers a_i (1 ≤ a_i ≤ {10}^{9}). If there are several solutions, output any of them. One can prove that if there is a solution, then there is also a solution in which all a_i are between 1 and 10^9.

If there are no solutions, print "NO".

Examples

Input


3
2 0
0 2
2 0


Output


YES
1 2 1 

Input


5
0 1
1 3
2 1
3 0
2 0


Output


YES
2 3 2 1 2
#include <bits/stdc++.h>
using namespace std;
struct Node {
  vector<long long int> ch;
  long long int c, val;
  Node() { c = 0; }
};
long long int solve(Node t[], long long int cur, long long int& ans) {
  long long int sum = 0;
  for (auto it : t[cur].ch) sum += solve(t, it, ans);
  if (sum < t[cur].c) ans = 0;
  return sum + 1;
}
void solve2(Node t[], long long int cur, set<long long int>& s) {
  long long int n = t[cur].c, l = 1, r = 1e+8;
  while (l < r) {
    long long int m = (l + r + 1) / 2;
    if (distance(s.begin(), s.lower_bound(m)) > n)
      r = m - 1;
    else
      l = m;
  }
  t[cur].val = l;
  s.erase(l);
  for (auto it : t[cur].ch) solve2(t, it, s);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, p, c, ans = 1;
  cin >> n;
  Node t[n + 1];
  for (long long int i = 1; i <= n; i++) {
    cin >> p >> c;
    t[p].ch.push_back(i);
    t[i].c = c;
  }
  solve(t, 0, ans);
  if (ans) {
    set<long long int> s;
    for (long long int i = 1; i <= n; i++) s.insert(i);
    solve2(t, t[0].ch[0], s);
    printf("YES\n");
    for (long long int i = 1; i <= n; i++) printf("%lld ", t[i].val);
  } else
    printf("NO");
  return 0;
}
