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 n, root, noch[2005], p[2005], c[2005], ans[2005], cur = 1;
vector<vector<long long int> > adj(2006);
vector<pair<long long int, long long int> > numof;
vector<long long int> ssol;
void dfss(long long int x) {
  for (auto it : adj[x]) {
    dfss(it);
  }
  ssol.push_back(ans[x]);
}
void dfs(long long int x) {
  long long int su = 0;
  for (auto it : adj[x]) {
    if (!noch[it]) {
      dfs(it);
      su += noch[it] + 1;
    }
  }
  noch[x] = su;
  numof.push_back(make_pair(x, su));
}
bool bo(pair<long long int, long long int> a,
        pair<long long int, long long int> b) {
  return a.second < b.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int x, y, z;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    adj[p[i]].push_back(i);
    if (p[i] == 0) {
      root = i;
    }
  }
  dfs(root);
  for (long long int i = 1; i <= n; i++) {
    if (c[i] > noch[i]) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES\n";
  sort(numof.begin(), numof.end(), bo);
  for (long long int i = 0; i < n; i++) {
    x = numof[i].first;
    y = c[x];
    ssol.clear();
    dfss(x);
    if (y + 1 == ssol.size()) {
      ans[x] = i + 1;
    } else {
      sort(ssol.begin(), ssol.end());
      ans[x] = ssol[y + 1];
      for (long long int j = 1; j <= n; j++) {
        if (j != x && ans[j] >= ans[x]) {
          ans[j]++;
        }
      }
    }
  }
  for (long long int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
