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>
struct Lbl {
  Lbl(int u, int v) : node(u), val(v) {}
  int node, val;
  bool operator<(const Lbl &other) const { return (node < other.node); }
};
const int MAXN = 2000;
int c[MAXN];
std::vector<int> tree[MAXN];
std::vector<Lbl> label(int u) {
  std::vector<Lbl> a;
  for (int v : tree[u]) {
    std::vector<Lbl> ans = label(v);
    if (ans.empty()) return {};
    int s = a.size();
    for (Lbl l : ans) a.emplace_back(l.node, l.val + s);
  }
  if ((int)a.size() < c[u]) return {};
  a.emplace_back(u, a.size() + 1);
  for (int i = a.size() - 1; i > c[u]; i--) std::swap(a[i].node, a[i - 1].node);
  return a;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cout.tie(nullptr);
  std::cin.tie(nullptr);
  int nodes;
  std::cin >> nodes;
  int root = 0;
  for (int iNode = 0; iNode < nodes; iNode++) {
    int p;
    std::cin >> p >> c[iNode];
    if (p == 0)
      root = iNode;
    else
      tree[p - 1].emplace_back(iNode);
  }
  std::vector<Lbl> a = label(root);
  if (a.empty())
    std::cout << "NO\n";
  else {
    std::sort(a.begin(), a.end());
    std::cout << "YES\n";
    for (Lbl l : a) std::cout << l.val << ' ';
    std::cout << '\n';
  }
  return 0;
}
