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;
long long int em(long long int x, long long int Mod, long long int n) {
  if (n == 0)
    return 1ll;
  else if (n % 2 == 0)
    return em((x % Mod * x % Mod) % Mod, Mod, n / 2) % Mod;
  else
    return (x % Mod * em((x % Mod * x % Mod) % Mod, Mod, (n - 1) / 2)) % Mod;
}
vector<long long int> v[500009], val[500009];
long long int c[500009];
void dfs(long long int x, long long int par) {
  for (long long int y : v[x]) {
    if (y == par) continue;
    dfs(y, x);
    for (long long int z : val[y]) val[x].push_back(z);
  }
  if (val[x].size() < c[x]) {
    cout << "NO\n";
    exit(0);
  }
  val[x].insert(val[x].begin() + c[x], x);
}
int32_t main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, j, k, l;
  cin >> n;
  for (long long int i = (long long int)1; i < (long long int)n + 1; i++) {
    cin >> k >> c[i];
    if (k == 0)
      j = i;
    else {
      v[k].push_back(i);
      v[i].push_back(k);
    }
  }
  dfs(j, 0);
  cout << "YES\n";
  long long int ans[n + 1];
  for (long long int i = (long long int)0; i < (long long int)n; i++)
    ans[val[j][i]] = i + 1;
  for (long long int i = (long long int)1; i < (long long int)n + 1; i++)
    cout << ans[i] << " ";
}
