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;
int ar[2004], sz[2004], ans[2004];
bool vis[2004];
vector<vector<int>> gr(2004), ord(2004);
void dfs(int v) {
  if (vis[v]) {
    return;
  }
  vis[v] = true;
  sz[v] = 1;
  for (int i : gr[v]) {
    if (vis[i]) {
      continue;
    }
    dfs(i);
    for (int j : ord[i]) {
      ord[v].push_back(j);
    }
    sz[v] += sz[i];
  }
  if (sz[v] - 1 < ar[v]) {
    cout << "No\n";
    exit(0);
  }
  ord[v].insert(ord[v].begin() + ar[v], v);
}
void solve() {
  int n, root, p, c;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p >> c;
    ar[i] = c;
    if (p != 0) {
      gr[i].push_back(p);
      gr[p].push_back(i);
    } else {
      root = i;
    }
  }
  dfs(root);
  c = 1;
  for (int i : ord[root]) {
    ans[i] = c++;
  }
  cout << "Yes\n";
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
