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 t, n, p[2005], c[2005], ans[2005], sum[2005];
vector<int> v[2005];
bool b[2005];
void pre(int first) {
  sum[first] = 1;
  for (int i = 0; i < v[first].size(); i++) {
    pre(v[first][i]);
    sum[first] += sum[v[first][i]];
  }
  return;
}
void f(int first, int l, int r) {
  if (sum[first] - 1 < c[first]) {
    printf("NO\n");
    exit(0);
  }
  int ct = c[first];
  for (int i = l; i <= r; i++) {
    if (!ct && !b[i]) {
      ct = i;
      break;
    }
    if (!b[i]) ct--;
  }
  ans[first] = ct;
  b[ct] = true;
  for (int i = 0; i < v[first].size(); i++) {
    f(v[first][i], 1, n);
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    v[p[i]].push_back(i);
  }
  int start = v[0][0];
  pre(start);
  f(start, 1, n);
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf(i == n ? "%d\n" : "%d ", ans[i]);
  }
  return 0;
}
