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 maxn = 2010;
int arr[maxn], sum[maxn], res[maxn];
vector<int> edges[maxn];
int lastOrder(int now, int &ptr) {
  if (edges[now].empty()) {
    if (sum[now]) {
      cout << "NO" << endl;
      exit(0);
    }
    arr[ptr++] = now;
    return 1;
  }
  int offset = 0, tmp;
  for (const auto child : edges[now]) offset += lastOrder(child, ptr);
  if (sum[now] > offset) {
    cout << "NO" << endl;
    exit(0);
  }
  int p = ptr;
  for (int i = 0; i < offset - sum[now]; ++i, --p) arr[p] = arr[p - 1];
  arr[p] = now;
  ++ptr;
  return offset + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, par, num;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> par >> num;
    edges[par].push_back(i);
    sum[i] = num;
  }
  int root = edges[0][0], ptr = 0;
  lastOrder(root, ptr);
  cout << "YES" << endl;
  for (int i = 0; i < n; ++i) res[arr[i]] = i + 1;
  for (int i = 1; i < n; ++i) cout << res[i] << " ";
  cout << res[n] << endl;
  return 0;
}
