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 long double pi = acos(-1);
const long long md = 1e9 + 7;
long long n, i, j, p, k[2501], t, in[2501], sz[2501], ans[2501], rt;
vector<long long> v, c[2501];
void dfs1(long long nom, long long prv) {
  t++;
  in[nom] = t;
  for (long long x : c[nom])
    if (x != prv) dfs1(x, nom);
  sz[nom] = t - in[nom];
}
void dfs2(long long nom, long long prv) {
  ans[nom] = v[k[nom]];
  v.erase(v.begin() + k[nom]);
  for (long long x : c[nom])
    if (x != prv) dfs2(x, nom);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> p >> k[i];
    if (p == 0) rt = i;
    c[p].push_back(i);
    v.push_back(i);
  }
  dfs1(rt, 0);
  for (i = 1; i <= n; i++) {
    if (k[i] > sz[i]) {
      cout << "NO" << '\n';
      return 0;
    }
  }
  dfs2(rt, 0);
  cout << "YES" << '\n';
  for (i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}
