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;
const int N = 2000, N_ = 2048;
int eo[N], *ej[N], cc[N], sz[N], aa[N];
int st[N_ * 2], n_;
void append(int i, int j) {
  int o = eo[i]++;
  if (o >= 2 && (o & o - 1) == 0)
    ej[i] = (int *)realloc(ej[i], (o << 1) * sizeof *ej[i]);
  ej[i][o] = j;
}
void update(int i, int a) {
  for (i += n_; i; i >>= 1) st[i] += a;
}
int query(int c) {
  c++;
  int i = 1;
  while (i < n_) {
    i <<= 1;
    if (st[i] < c) {
      c -= st[i];
      i |= 1;
    }
  }
  update(i - n_, -1);
  return i - n_;
}
int dfs(int p, int i) {
  aa[i] = query(cc[i]);
  int k = 1;
  for (int o = eo[i]; o--;) {
    int j = ej[i][o];
    if (j != p) k += dfs(i, j);
  }
  return sz[i] = k;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) ej[i] = (int *)malloc(2 * sizeof *ej[i]);
  int r = -1;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> cc[i], p--;
    if (p != -1)
      append(i, p), append(p, i);
    else
      r = i;
  }
  n_ = 1;
  while (n_ < n) n_ <<= 1;
  for (int a = 0; a < n; a++) update(a, 1);
  dfs(-1, r);
  for (int i = 0; i < n; i++)
    if (cc[i] > sz[i] - 1) {
      cout << "NO\n";
      return 0;
    }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << aa[i] + 1 << ' ';
  cout << '\n';
  return 0;
}
