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 = 2000 + 3;
int n, c[MAXN], sz[MAXN], r, z[MAXN];
vector<int> child[MAXN], tmp[MAXN];
int sfs(int v) {
  sz[v] = 1;
  for (auto u : child[v]) sz[v] += sfs(u);
  return sz[v];
}
void dfs(int v) {
  if (child[v].empty()) return tmp[v].push_back(v), void(0);
  for (auto u : child[v]) dfs(u);
  if (tmp[v].size() == c[v]) tmp[v].push_back(v);
  for (auto u : child[v])
    for (auto j : tmp[u]) {
      tmp[v].push_back(j);
      if (tmp[v].size() == c[v]) tmp[v].push_back(v);
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i];
    if (p != 0)
      child[--p].push_back(i);
    else
      r = i;
  }
  sfs(r);
  for (int i = 0; i < n; i++)
    if (sz[i] <= c[i]) return cout << "NO", 0;
  dfs(r);
  cout << "YES\n";
  for (int i = 0; i < n; i++) z[tmp[r][i]] = i + 1;
  for (int i = 0; i < n; i++) cout << z[i] << ' ';
  return 0;
}
