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;
#pragma GCC optimize("unroll-loops", "no-stack-protector")
#pragma GCC optimize("Ofast", "fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
int n;
vector<vector<int> > g;
vector<int> c;
vector<int> a;
vector<int> siz;
vector<int> p;
int cnt = 1;
void precalc(int v) {
  siz[v] = 1;
  for (auto to : g[v]) {
    precalc(to);
    siz[v] += siz[to];
  }
  if (c[v] >= siz[v]) {
    cout << "NO";
    exit(0);
  }
}
vector<pair<int, int> > buf;
void get(int v) {
  buf.push_back({a[v], v});
  for (auto to : g[v]) {
    get(to);
  }
}
void f(int v) {
  for (auto to : g[v]) {
    f(to);
  }
  if (g[v].size() == 0) {
    a[v] = cnt;
    cnt++;
    return;
  }
  if (c[v] == siz[v] - 1) {
    a[v] = cnt;
    cnt++;
    return;
  }
  buf.clear();
  for (auto to : g[v]) {
    get(to);
  }
  sort(buf.begin(), buf.end());
  a[v] = buf[c[v]].first;
  for (int i = c[v]; i < buf.size(); i++) {
    a[buf[i].second]++;
  }
  cnt++;
  return;
}
int main() {
  cin >> n;
  p.resize(n);
  siz.resize(n);
  a.resize(n);
  c.resize(n);
  g.resize(n);
  int r = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x >> c[i];
    x--;
    if (x == -1) {
      r = i;
      continue;
    }
    p[i] = x;
    g[x].push_back(i);
  }
  precalc(r);
  cout << "YES\n";
  f(r);
  for (int i = 0; i < n; i++) cout << a[i] << " ";
}
