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;
void itval(istream_iterator<string> it) {}
template <typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  itval(++it, args...);
}
const long long int MOD = 1e9 + 9;
template <typename T>
inline void print(T x) {
  cout << x << "\n";
}
template <typename T>
inline void printvec(T x) {
  for (auto a : x) cout << a << ' ';
  cout << '\n';
}
struct custom {
  bool operator()(
      const pair<pair<long long int, long long int>, long long int> &p1,
      const pair<pair<long long int, long long int>, long long int> &p2) const {
    if (p1.first.first == p2.first.first) {
      if (p1.first.second == p2.first.second) return p1.second < p2.second;
      return p1.first.second < p2.first.second;
    }
    return p1.first.first < p2.first.first;
  }
};
long long int get_pow(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
const long long int N = 2e3 + 10, inf = 4e18;
std::vector<vector<int> > adj(N);
vector<int> c(N), p(N);
vector<int> dfs(int s) {
  vector<int> res;
  for (auto x : adj[s]) {
    vector<int> z = dfs(x);
    res.insert(res.end(), z.begin(), z.end());
  }
  if (res.size() < c[s]) {
    cout << "NO\n";
    exit(0);
  }
  res.insert(res.begin() + c[s], s);
  return res;
}
void solve() {
  int n, x, root;
  cin >> n;
  for (long long int i = (long long int)0; i < (long long int)(n); i++) {
    cin >> p[i] >> c[i];
    p[i]--;
    if (p[i] != -1)
      adj[p[i]].push_back(i);
    else
      root = i;
  }
  vector<int> dum = dfs(root);
  vector<int> ans(n);
  for (long long int i = (long long int)0; i < (long long int)(n); i++) {
    ans[dum[i]] = i + 1;
  }
  cout << "YES\n";
  printvec(ans);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  clock_t z = clock();
  for (long long int tes = (long long int)0; tes < (long long int)(test);
       tes++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}
