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;
const long long mod = 1e9 + 7;
template <class T>
inline T read() {
  T x = 0, f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  };
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + s - 48;
    s = getchar();
  };
  return x * f;
}
template <typename T>
inline void wri(T x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  };
  if (x / 10) wri(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void print(T x, char s) {
  wri(x);
  putchar(s);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
const int N = 2010;
vector<int> node[N];
int c[N], ans[N];
vector<int> dfs(int u) {
  vector<int> rank;
  for (auto v : node[u]) {
    vector<int> temp = dfs(v);
    rank.insert(rank.end(), temp.begin(), temp.end());
  }
  if (c[u] > rank.size()) {
    cout << "NO";
    exit(0);
  }
  rank.insert(rank.begin() + c[u], u);
  return rank;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, root;
  cin >> n;
  for (register int i = (1); i < (n + 1); i++) {
    int fa;
    cin >> fa >> c[i];
    if (fa == 0)
      root = i;
    else
      node[fa].push_back(i);
  }
  vector<int> rank = dfs(root);
  for (register int i = (0); i < (rank.size()); i++) {
    ans[rank[i]] = i + 1;
  }
  cout << "YES" << endl;
  for (register int i = (1); i < (n + 1); i++) cout << ans[i] << " ";
  return 0;
}
