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;
long long bp(long long x, long long y) {
  long long ans = 1;
  x %= mod;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long mod_inverse(long long x) { return bp(x, mod - 2); }
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long max(long long a, long long b) {
  if (a < b) swap(a, b);
  return a;
}
long long min(long long a, long long b) {
  if (a > b) swap(a, b);
  return a;
}
long long countSetBits(long long n) {
  long long count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
vector<long long> adj[2005];
long long tree[2005][2], val[2005];
long long root = -1;
vector<pair<long long, long long>> dfs(long long root, long long parent) {
  vector<pair<long long, long long>> res, temp;
  long long shift = 0;
  for (auto i : adj[root]) {
    if (i == parent) continue;
    temp = dfs(i, root);
    if (temp.size() == 0) return temp;
    for (auto j : temp) res.push_back({j.first, j.second + shift});
    shift = res.back().second;
  }
  if (tree[root][1] > res.size()) return vector<pair<long long, long long>>();
  res.insert(res.begin() + tree[root][1],
             pair<long long, long long>(root, tree[root][1] + 1));
  for (long long i = tree[root][1] + 1; i < res.size(); i++) res[i].second++;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> tree[i][0] >> tree[i][1];
    adj[i].push_back(tree[i][0]), adj[tree[i][0]].push_back(i);
    if (tree[i][0] == 0) root = i;
  }
  vector<pair<long long, long long>> res = dfs(root, 0);
  if (res.size() == 0) {
    cout << "NO";
    return;
  }
  cout << "YES" << endl;
  for (auto i : res) val[i.first] = i.second;
  for (long long i = 1; i <= n; i++) cout << val[i] << " ";
}
signed main() { solve(); }
