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 = 2005;
int N, M, K;
vector<int> G[n];
int C[n], A[n];
bool O[n];
bool pos = true;
void DFS(int v) {
  int c = C[v] + 1;
  for (int i = 0; i < N; ++i) {
    if (!O[i]) A[v] = i, c--;
    if (c == 0) break;
  }
  if (c != 0) pos = false;
  O[A[v]] = true;
  for (int c : G[v]) {
    DFS(c);
  }
  for (int i = 0; i < A[v]; ++i) {
    pos = pos && O[i];
  }
}
int main() {
  cin >> N;
  int root = -1;
  for (int i = 0; i < N; ++i) {
    int p;
    cin >> p >> C[i];
    p--;
    if (p == -1)
      root = i;
    else
      G[p].push_back(i);
  }
  DFS(root);
  if (pos) {
    cout << "YES\n";
    for (int i = 0; i < N; ++i) {
      cout << A[i] + 1 << " ";
    }
    cout << endl;
  } else {
    cout << "NO\n";
  }
  return 0;
}
