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;
template <class T>
ostream& operator<<(ostream& stream, const vector<T> v) {
  stream << "[ ";
  for (int i = 0; i < (int)v.size(); i++) stream << v[i] << " ";
  stream << "]";
  return stream;
}
long long fpow(long long x, long long p, long long m) {
  long long r = 1;
  for (; p; p >>= 1) {
    if (p & 1) r = r * x % m;
    x = x * x % m;
  }
  return r;
}
long long inv(long long a, long long b) {
  return a < 2 ? a : ((a - inv(b % a, a)) * b + 1) / a % b;
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int N, p[2010], c[2010], K, a[2010];
vector<int> g[2010];
void collect(int nod, vector<pair<long long, long long> >& arr) {
  arr.push_back({a[nod], nod});
  for (int nxt : g[nod]) collect(nxt, arr);
}
void dfs(int nod) {
  if (g[nod].size() == 0) {
    if (c[nod] != 0) {
      cout << "NO\n";
      exit(0);
    }
    a[nod] = 100000;
    return;
  }
  vector<pair<long long, long long> > arr;
  for (int nxt : g[nod]) {
    dfs(nxt);
    collect(nxt, arr);
  }
  sort((arr).begin(), (arr).end());
  int sz = arr.size();
  if (sz < c[nod]) {
    cout << "NO\n";
    exit(0);
  }
  if (c[nod] == sz)
    a[nod] = arr[sz - 1].first + 1;
  else {
    a[nod] = arr[c[nod]].first;
    int i;
    for (i = 0; i < c[nod]; i++) a[arr[i].second]--;
    for (i = c[nod]; i < sz; i++) a[arr[i].second]++;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  int i, r;
  for (i = 1; i <= N; i++) {
    cin >> p[i] >> c[i];
    if (p[i] == 0)
      r = i;
    else
      g[p[i]].push_back(i);
  }
  dfs(r);
  cout << "YES\n";
  for (i = 1; i <= N; i++) cout << a[i] << " ";
  return 0;
}
