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;
long long n, x, a[2005], par[2005], dp[2005], c[2005];
bitset<2005> b;
bool f;
vector<int> v[2005];
int bruteforce(long long x) {
  int jum = 0;
  for (int i = 1; i <= n; i++) {
    if (!b[i]) {
      jum++;
      if (jum == x + 1) {
        return i;
      }
    }
  }
  return -1;
}
void dfs(long long x) {
  int tmp = bruteforce(c[x]);
  if (tmp == -1) {
    f = true;
    return;
  }
  b[tmp] = true, a[x] = tmp;
  for (int i = 0; i < v[x].size(); i++) {
    dfs(v[x][i]);
    dp[x] += dp[v[x][i]];
  }
  if (dp[x] < c[x]) {
    f = true;
    return;
  }
  dp[x]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> par[i] >> c[i];
    v[par[i]].push_back(i);
  }
  dfs(v[0][0]);
  if (f) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) {
      cout << a[i] << " ";
    }
    cout << '\n';
  }
}
