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 inf = 0x3f3f3f3f;
const int N = 2010;
const int mod = 1e9 + 7;
vector<int> xiao[N];
int arr[N], ans[N], vis[N], n, flag;
int dfs(int rt) {
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      sum++;
      if (sum == arr[rt] + 1) {
        ans[rt] = i;
        vis[i] = 1;
        break;
      }
    }
  }
  int num = 0;
  for (int i = 0; i < xiao[rt].size(); i++) num += dfs(xiao[rt][i]);
  if (num < arr[rt]) flag = 1;
  return num + 1;
}
int main() {
  int rt;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int fa;
    scanf("%d%d", &fa, &arr[i]);
    if (fa == 0)
      rt = i;
    else
      xiao[fa].push_back(i);
  }
  dfs(rt);
  if (flag)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}
