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 long long MOD = 1e9 + 7;
const long long MOD2 = 998244353;
const long long N = 2e5 + 5;
const long double pi = 3.14159265359;
long long n, pa, c[N], rt, a[N], con, vis[N], sz[N];
vector<long long> v[N];
void DFS(long long nd) {
  long long ti = 0;
  sz[nd] = 1;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) ti++;
    if (ti == c[nd] + 1) {
      a[nd] = i;
      vis[a[nd]] = 1;
      break;
    }
  }
  if (a[nd] == -1) con = 1;
  for (auto i : v[nd]) {
    DFS(i);
    sz[nd] += sz[i];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(a, -1, sizeof(a));
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> pa >> c[i];
    if (pa == 0)
      rt = i;
    else
      v[pa].push_back(i);
  }
  DFS(rt);
  for (long long i = 1; i <= n; i++) {
    if (sz[i] - 1 < c[i]) con = 1;
  }
  if (con) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (long long i = 1; i <= n; i++) cout << a[i] << (i == n ? "\n" : " ");
  return 0;
}
