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;
using ll = long long;
using ii = pair<int, int>;
constexpr int MAXN = 5 + 2000;
int c[MAXN];
vector<int> g[MAXN];
bool sol = true;
int mincnt = 1;
set<ii> solve(int u) {
  set<ii> res;
  int cumu = 0;
  for (int v : g[u]) {
    const set<ii>& r = solve(v);
    for (const auto& x : r) {
      res.insert({cumu + x.first, x.second});
    }
    cumu += (int)r.size();
  }
  if (c[u] > (int)res.size()) {
    sol = false;
  } else {
    set<ii>::iterator it = res.begin();
    advance(it, c[u]);
    res.insert({1 + (it != res.begin() ? prev(it)->first : 0), u});
    while (it != res.end()) {
      const auto p = *it;
      while (it != res.end() && it->first == p.first) it = next(it);
      res.insert({1 + p.first, p.second});
      res.erase(p);
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, r = -1;
  cin >> n;
  for (int i = (int)0; i < (int)n; ++i) {
    int a;
    cin >> a >> c[i];
    --a;
    if (a >= 0) {
      g[a].push_back(i);
    } else {
      assert(r == -1);
      r = i;
    }
  }
  assert(~r);
  const set<ii>& res = solve(r);
  if (!sol) {
    cout << "NO" << '\n';
  } else {
    vector<int> ans(n);
    for (const auto& x : res) {
      ans[x.second] = x.first;
    }
    cout << "YES\n";
    for (int x : ans) {
      cout << x << ' ';
    }
    cout << '\n';
  }
  return 0;
}
