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 Maxn = 2005;
bool tk[Maxn];
int n;
vector<int> neigh[Maxn];
int c[Maxn], cnt[Maxn];
int root;
int res[Maxn];
int Get(int ind) {
  for (int i = 1; i <= n; i++)
    if (!tk[i])
      if (ind == 1) {
        tk[i] = true;
        return i;
      } else
        ind--;
  return -1;
}
void Count(int v) {
  cnt[v] = 1;
  for (int i = 0; i < neigh[v].size(); i++) {
    int u = neigh[v][i];
    Count(u);
    cnt[v] += cnt[u];
  }
}
bool Solve(int v) {
  if (c[v] + 1 > cnt[v]) return false;
  res[v] = Get(c[v] + 1);
  for (int i = 0; i < neigh[v].size(); i++) {
    int u = neigh[v][i];
    if (!Solve(u)) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d %d", &p, &c[i]);
    if (p == 0)
      root = i;
    else
      neigh[p].push_back(i);
  }
  Count(root);
  if (!Solve(root))
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= n; i++)
      printf("%d%c", res[i], i + 1 <= n ? ' ' : '\n');
  }
  return 0;
}
