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;
set<long long> st;
vector<long long> adj[2001];
long long p[2001], c[2001], sz[2001], ans[2001];
long long dfs(long long i = 1, long long pr = 0) {
  sz[i] = 1;
  for (auto j : adj[i]) {
    if (j != pr) {
      sz[i] += dfs(j, i);
    }
  }
  return sz[i];
}
void func(long long i = 1, long long pr = 0) {
  long long x = c[i];
  auto itr = st.begin();
  while (x--) {
    itr++;
  }
  ans[i] = *itr;
  st.erase(itr);
  for (auto j : adj[i]) {
    if (j != pr) {
      func(j, i);
    }
  }
}
int main() {
  long long n, i, parent;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    if (p[i] == 0) {
      parent = i;
    }
    adj[p[i]].push_back(i);
    st.insert(i);
  }
  dfs(parent);
  for (int i = 1; i <= n; i++) {
    if (sz[i] <= c[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  func(parent);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
