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 int;
vector<int> ad[2005];
int ara[2005], sz[2005], ans_ara[2005];
set<int> st;
void dfs(int src, int pre) {
  sz[src] = 1;
  for (int v : ad[src]) {
    if (v == pre) continue;
    dfs(v, src);
    sz[src] += sz[v];
  }
}
void dfs2(int src, int pre) {
  set<int>::iterator it;
  int cnt = 0;
  for (it = st.begin(); it != st.end(); it++) {
    if (cnt == ara[src]) {
      ans_ara[src] = *it;
      st.erase(*it);
      break;
    }
    ++cnt;
  }
  for (int v : ad[src]) {
    if (v == pre) continue;
    dfs2(v, src);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, i, j, root, p;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> p >> ara[i];
    if (!p) root = i;
    if (p) {
      ad[i].push_back(p);
      ad[p].push_back(i);
    }
  }
  dfs(root, 0);
  for (i = 1; i <= n; i++) {
    if (ara[i] > sz[i] - 1) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  for (i = 1; i <= n; i++) st.insert(i);
  dfs2(root, 0);
  for (i = 1; i <= n; i++) cout << ans_ara[i] << ' ';
  cout << endl;
  return 0;
}
