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 = 1000000007;
long double PI = 4 * atan(1);
int n, p[2001], c[2001], size[2001], r, flag, ans[2001];
vector<int> v[2001];
int dfs(int i) {
  for (int a : v[i]) {
    size[i] += dfs(a);
  }
  size[i]++;
  return size[i];
}
void dfs2(int i, vector<int> cur) {
  if (c[i] >= cur.size()) {
    flag = 1;
    return;
  }
  ans[i] = cur[c[i]], cur.erase(cur.begin() + c[i]);
  int cnt = 0;
  for (int a : v[i]) {
    vector<int> V;
    for (int j = 0; j < size[a]; j++) V.push_back(cur[cnt++]);
    dfs2(a, V);
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> p[i] >> c[i];
    if (p[i] == 0) r = i;
    v[p[i]].push_back(i);
  }
  dfs(r);
  vector<int> cur;
  for (int i = 1; i < n + 1; i++) cur.push_back(i);
  dfs2(r, cur);
  if (flag) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (int i = 1; i < n + 1; i++) cout << ans[i] << " ";
  return 0;
}
