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 namespace std;
const long long int mod = 1e9 + 7;
const long long int MAX = LLONG_MAX;
const long long int MIN = LLONG_MIN;
vector<long long int> adj[2010];
long long int vis[2010] = {0};
long long int cnt[2010] = {0};
long long int vert[2010] = {0};
vector<long long int> order[2010];
void add(vector<long long int> &a, vector<long long int> &b) {
  for (auto i : b) a.push_back(i);
}
void dfs(long long int v) {
  vis[v] = 1;
  vert[v] = 1;
  for (long long int i = 0; i < adj[v].size(); i++) {
    long long int child = adj[v][i];
    if (vis[child] == 0) {
      dfs(child);
      add(order[v], order[child]);
      vert[v] += vert[child];
    }
  }
  if (cnt[v] > vert[v] - 1) {
    cout << "NO\n";
    exit(0);
  }
  order[v].insert(order[v].begin() + cnt[v], v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n;
  cin >> n;
  long long int root;
  for (long long int i = 1; i < n + 1; i++) {
    long long int x, y;
    cin >> x >> y;
    cnt[i] = y;
    if (x == 0)
      root = i;
    else {
      adj[x].push_back(i);
      adj[i].push_back(x);
    }
  }
  dfs(root);
  long long int ans[2010];
  for (int i = 0; i < n; i++) {
    ans[order[root][i]] = i + 1;
  }
  cout << "YES\n";
  for (long long int i = 1; i < n + 1; i++) cout << ans[i] << " ";
  return 0;
}
