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 int mod = 1e9 + 7;
int sz[2005];
int nd[2005];
vector<int> g[2005];
int ans[2005];
int dfs1(int now) {
  sz[now] = 1;
  for (auto i : g[now]) {
    sz[now] += dfs1(i);
  }
  return sz[now];
}
void dfs2(int now, vector<int> v) {
  ans[now] = v[nd[now]];
  int ptr = 0;
  for (auto i : g[now]) {
    vector<int> vv;
    while ((int)vv.size() < sz[i]) {
      if (ptr == nd[now])
        ;
      else
        vv.push_back(v[ptr]);
      ptr++;
    }
    dfs2(i, vv);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  while (cin >> n) {
    int rt;
    for (int i = 1; i < n + 1; i++) {
      int add;
      cin >> add;
      if (add == 0) rt = i;
      g[add].push_back(i);
      cin >> nd[i];
    }
    dfs1(rt);
    for (int i = 1; i < n + 1; i++) {
      if (nd[i] >= sz[i]) {
        cout << "NO\n";
        exit(0);
      }
    }
    vector<int> v;
    for (int i = 1; i < n + 1; i++) {
      v.push_back(i);
    }
    dfs2(rt, v);
    cout << "YES\n";
    for (int i = 1; i < n + 1; i++) {
      cout << ans[i] << ' ';
    }
    cout << '\n';
  }
}
