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 <typename T>
using second = set<T>;
const int Max = 3e5;
const int Mx = 2e6;
multiset<string> avalableStrings;
string str[1500];
int evenCnt = 0;
int oddCnt = 0;
int cnt[2010];
int par[2010];
int deg[2010];
int val[2010];
vector<int> N[2010];
const int Dist = 4000;
int lastUsingLeaf = 1;
multiset<pair<int, int> > usingValues[2010];
int childCnt[2010];
int DFS1(int root) {
  int sum = 0;
  for (auto v : N[root]) {
    sum += DFS1(v);
  }
  childCnt[root] = sum + 1;
  return sum + 1;
}
void DFS(int root) {
  for (auto n : N[root]) {
    DFS(n);
    for (auto v : usingValues[n]) usingValues[root].insert(v);
  }
  if (N[root].empty()) {
    val[root] = lastUsingLeaf;
    usingValues[root].insert({val[root], root});
    lastUsingLeaf += Dist;
  } else {
    auto itr = usingValues[root].begin();
    if (cnt[root] == 0) {
      val[root] = itr->first - 1;
      usingValues[root].insert({val[root], root});
    } else {
      for (int i = 0; i < cnt[root] - 1; i++) {
        ++itr;
      }
      val[root] = itr->first + 1;
      int trg = val[root];
      auto ub = usingValues[root].begin();
      vector<pair<int, int> > edittedList;
      while ((ub = usingValues[root].upper_bound({trg, -1})) !=
                 usingValues[root].end() &&
             ub->first == trg) {
        edittedList.push_back({ub->first, ub->second});
        val[ub->second] = ub->first + 1;
        trg = val[ub->second];
      }
      for (auto v : edittedList) {
        usingValues[root].erase(v);
        usingValues[root].insert({val[v.second], v.second});
      }
      usingValues[root].insert({val[root], root});
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> par[i + 1] >> cnt[i + 1];
    N[par[i + 1]].push_back(i + 1);
  }
  DFS1(0);
  for (int i = 0; i <= n; i++) {
    if (childCnt[i] <= cnt[i]) {
      cout << "NO\n";
      return 0;
    }
  }
  int lastUsedForLeaf = 0;
  DFS(0);
  int mn = *min_element(val, val + n + 1);
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) {
    cout << val[i] + (1 - mn) << " ";
  }
}
