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 n;
int root;
vector<int> children[2005];
int ans[2005];
int num[2005];
int cc[2005];
void add(int v, int a) {
  ans[v] += a;
  for (auto i : children[v]) add(i, a);
}
void check(int v, int i) {
  if (ans[v] > cc[i]) ans[v]++;
  for (auto k : children[v]) check(k, i);
}
void getAns(int v) {
  num[v] = 1;
  int prev = 0;
  for (auto i : children[v]) {
    getAns(i);
    add(i, prev);
    prev += num[i];
    num[v] += num[i];
  }
  if (cc[v] >= num[v]) {
    cout << "NO" << endl;
    exit(0);
  }
  check(v, v);
  if (children[v].size() == 0) {
    ans[v] = 1;
  } else {
    ans[v] = cc[v] + 1;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p, c;
    cin >> p >> c;
    --p;
    if (p == -1) {
      root = i;
    } else
      children[p].push_back(i);
    cc[i] = c;
  }
  getAns(root);
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
