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 = 2020;
long long tl[N], ans[N], c[N];
vector<long long> val(2 * N), v[N];
void dfs(long long ver) {
  ans[ver] = val[c[ver]];
  val.erase(val.begin() + c[ver]);
  for (auto it : v[ver]) {
    dfs(it);
    tl[ver] += tl[it] + 1;
  }
}
int main() {
  std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, root, x;
  cin >> n;
  for (int i = int(1); i <= int(n); i++) val[i - 1] = i;
  for (int i = int(1); i <= int(n); i++) {
    cin >> x >> c[i];
    v[x].push_back(i);
    if (x == 0) root = i;
  }
  dfs(root);
  for (int i = int(1); i <= int(n); i++) {
    if (c[i] > tl[i]) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES"
       << "\n";
  for (int i = int(1); i <= int(n); i++) cout << ans[i] << " ";
}
