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 int N = 3e3;
int p[N], c[N];
vector<int> g[N];
bool FAIL = false;
inline int max(int a, int b) { return a > b ? a : b; }
vector<int> f(int v, int pr = -1) {
  vector<int> res, w;
  bool done = false;
  for (int u : g[v]) {
    if (u == pr) continue;
    w = f(u, v);
    if (w.size() <= c[v] || done) {
      if (!done) c[v] -= w.size();
      for (int x : w) res.push_back(x);
      continue;
    }
    done = true;
    for (int i = 0; i < c[v]; ++i) res.push_back(w[i]);
    res.push_back(v);
    for (int i = c[v]; i < w.size(); ++i) res.push_back(w[i]);
    c[v] = 0;
  }
  if (!done) res.push_back(v);
  if (c[v]) FAIL = true;
  return res;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i] >> c[i], --p[i];
    if (p[i] + 1) g[i].push_back(p[i]), g[p[i]].push_back(i);
  }
  int root;
  for (int i = 0; i < n; ++i)
    if (p[i] == -1) root = i;
  vector<int> a = f(root, -1);
  if (FAIL) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  int num[n];
  for (int i = 0; i < a.size(); ++i) num[a[i]] = i + 1;
  for (int i = 0; i < n; ++i) cout << num[i] << ' ';
  return 0;
}
