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;
int a[200000];
int c[200000];
vector<int> e[200000];
int p[200000];
int b[200000];
int pp;
bool vv;
int dfs(int k) {
  int re = 0;
  for (int i = 0; i < int(e[k].size()); i++) re += dfs(e[k][i]);
  if (a[k] > re) vv = false;
  b[k] = re + 1;
  return (re + 1);
}
vector<pair<int, int> > dfs2(int k) {
  vector<pair<int, int> > q;
  vector<pair<int, int> > qq;
  if (b[k] == 1) {
    c[k] = 1;
    qq.push_back(make_pair(c[k], k));
    return qq;
  }
  for (int i = 0; i < int(e[k].size()); i++) {
    q = dfs2(e[k][i]);
    for (int i = 0; i < int(q.size()); i++) {
      qq.push_back(q[i]);
    }
  }
  sort(qq.begin(), qq.end());
  c[k] = a[k] + 1;
  for (int i = 0; i < int(qq.size()); i++) {
    if (i < a[k]) {
      qq[i].first = i + 1;
      c[qq[i].second] = qq[i].first;
    } else {
      qq[i].first = i + 2;
      c[qq[i].second] = qq[i].first;
    }
  }
  qq.push_back(make_pair(c[k], k));
  return qq;
}
int main() {
  for (int i = 0; i < int(200000); i++) a[i] = 0;
  int n, k;
  int x, y, z;
  scanf("%d", &n);
  for (int i = 0; i < int(n); i++) {
    scanf("%d %d", &p[i], &a[i]);
    p[i]--;
    if (p[i] >= 0)
      e[p[i]].push_back(i);
    else
      pp = i;
  }
  vv = true;
  dfs(pp);
  if (!vv) {
    printf("NO\n");
    return 0;
  }
  dfs2(pp);
  printf("YES\n");
  for (int i = 0; i < int(n); i++) {
    printf("%d", c[i]);
    if (i < (n - 1))
      printf(" ");
    else
      printf("\n");
  }
  return 0;
}
