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;
vector<int> E[2010];
int w[2010], root, n, Num[2010], ord[2010], cnt, vis[2010], Res[2010],
    Rank[2010], Ed[2010], inp[2010];
void DFS(int a, int pp) {
  Num[a] = ++cnt;
  ord[cnt] = a;
  for (auto &x : E[a]) {
    DFS(x, a);
  }
  Ed[a] = cnt;
}
int main() {
  int i, a, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (a) {
      E[a].push_back(i);
    } else {
      root = i;
    }
    scanf("%d", &w[i]);
    inp[i] = w[i];
  }
  DFS(root, 0);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (!vis[ord[j]] && !w[ord[j]]) break;
    }
    if (j == n + 1) {
      puts("NO");
      return 0;
    }
    int x = ord[j];
    vis[x] = 1;
    Res[i] = x;
    Rank[x] = i;
    for (j = 1; j <= n; j++) {
      if (Num[j] < Num[x] && Num[x] <= Ed[j] && !vis[j]) {
        w[j]--;
        if (w[j] < 0) {
          puts("NO");
          return 0;
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    int c = 0;
    for (j = 1; j <= n; j++) {
      if (Num[i] < Num[j] && Num[j] <= Ed[i]) {
        if (Rank[i] > Rank[j]) c++;
      }
    }
    if (c != inp[i]) {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  for (i = 1; i <= n; i++) printf("%d ", Rank[i]);
}
