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 {
  int no;
  int visit;
  list<int> lst;
  list<int> edge;
  int c;
  int a;
};
void modifieddfs(node a[], int index, int parent) {
  if (a[index].visit == 1) {
    return;
  }
  a[index].visit = 1;
  for (auto it = a[index].edge.begin(); it != a[index].edge.end(); it++) {
    modifieddfs(a, *it, index);
  }
  int ctr = 0;
  if (ctr == a[index].c) {
    ctr++;
    a[index].lst.push_back(index);
  }
  for (auto it = a[index].edge.begin(); it != a[index].edge.end(); it++) {
    if (*it == parent) {
      continue;
    }
    for (auto it2 = a[*it].lst.begin(); it2 != a[*it].lst.end(); it2++) {
      ctr++;
      a[index].lst.push_back(*it2);
      if (ctr == a[index].c) {
        ctr++;
        a[index].lst.push_back(index);
      }
    }
  }
  if (ctr < a[index].c + 1) {
    a[index].a = -1;
  }
}
int main() {
  int n;
  cin >> n;
  node a[n];
  int root;
  for (int i = 0; i < n; i++) {
    int p, c;
    cin >> p >> c;
    a[i].c = c;
    if (p == 0) {
      root = i;
    } else {
      a[i].edge.push_back(p - 1);
      a[p - 1].edge.push_back(i);
    }
    a[i].no = i;
    a[i].visit = 0;
    a[i].a = 0;
  }
  modifieddfs(a, root, -1);
  int assignment = 1;
  for (int i = 0; i < n; i++) {
    if (a[i].a == -1) {
      cout << "NO";
      return 0;
    }
  }
  for (auto it = a[root].lst.begin(); it != a[root].lst.end(); it++) {
    a[*it].a = assignment++;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << a[i].a << " ";
  }
  return 0;
}
