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 N = 2e3 + 7;
int n;
vector<int> vnode[N], child[N];
int par[N], c[N], root, val[N];
bool mungkin = 1;
void sol(int node) {
  if (!mungkin) return;
  vector<int> temp;
  for (auto dest : child[node]) {
    sol(dest);
    for (auto x : vnode[dest]) {
      temp.push_back(x);
    }
  }
  int len = temp.size();
  if (len < c[node]) {
    mungkin = 0;
    return;
  }
  for (int i = 0; i < len; i++) {
    if (i == c[node]) {
      vnode[node].push_back(node);
    }
    vnode[node].push_back(temp[i]);
  }
  if (c[node] == len) {
    vnode[node].push_back(node);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> par[i] >> c[i];
    if (par[i] == 0) {
      root = i;
      continue;
    }
    child[par[i]].push_back(i);
  }
  sol(root);
  if (!mungkin) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  for (int i = 0; i < n; i++) {
    val[vnode[root][i]] = i + 1;
  }
  for (int i = 1; i <= n; i++) {
    cout << val[i] << " ";
  }
  cout << '\n';
  return 0;
}
