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 mx = 105;
const long long maxn = 1e5 + 1;
const long long mod = 1000000007;
vector<long long> ans;
vector<vector<int>> adj;
vector<long long> c;
bool pos;
vector<long long> dfs(long long v, long long p) {
  long long i, j, k;
  vector<long long> te;
  for (i = 0; i < (int)adj[v].size(); i++) {
    if (adj[v][i] != p) {
      vector<long long> te1 = dfs(adj[v][i], v);
      te.insert(te.begin(), te1.begin(), te1.end());
    }
  }
  if (c[v] > (int)te.size()) {
    cout << "NO\n";
    exit(0);
  }
  te.insert(te.begin() + c[v], v);
  return te;
}
void solve() {
  long long n, i, j, k, p;
  cin >> n;
  pos = true;
  adj.clear();
  adj.resize(n + 1);
  c.clear();
  c.resize(n + 1);
  long long r;
  for (i = 1; i <= n; i++) {
    cin >> p >> c[i];
    if (p != 0) {
      adj[i].push_back(p);
      adj[p].push_back(i);
    } else {
      r = i;
    }
  }
  long long res[n + 1];
  ans = dfs(r, -1);
  if (pos) {
    for (i = 0; i < n; i++) {
      res[ans[i]] = i + 1;
    }
    cout << "YES\n";
    for (i = 1; i <= n; i++) cout << res[i] << " ";
    cout << "\n";
  } else {
    cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  long long n, i, j, k;
  while (t--) {
    solve();
  }
}
