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>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
const int N = 2007;
int n, P[N], C[N], root, sz[N], val[N];
vector<int> G[N];
vector<int> ord;
int dfs(int v) {
  for (auto u : G[v]) sz[v] += dfs(u);
  ord.push_back(v);
  return ++sz[v];
}
void add(int v, int x, int y) {
  if (val[v] > x) val[v] += y;
  for (auto u : G[v]) add(u, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> P[i] >> C[i];
    if (P[i] == 0)
      root = i;
    else
      G[P[i]].push_back(i);
  }
  dfs(root);
  for (int i = 1; i <= n; i++) {
    if (!(0 <= C[i] && C[i] < sz[i])) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << "YES\n";
  for (auto v : ord) {
    int sm = 0;
    for (auto u : G[v]) {
      add(u, 0, sm);
      sm += sz[u];
    }
    val[v] = C[v] + 1;
    for (auto u : G[v]) {
      add(u, C[v], 1);
    }
  }
  for (int i = 1; i <= n; i++) cout << val[i] << " ";
  cout << "\n";
  return 0;
}
