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;
struct node {
  int to;
  int c;
};
vector<node> v[2005];
int vis[2005], f = 0;
int pos[2005];
vector<int> dfs(int u, int fa) {
  vector<int> a;
  for (int i = 0; i < v[u].size(); i++) {
    int to = v[u][i].to;
    if (to != fa) {
      vector<int> b = dfs(to, u);
      for (int i = 0; i < b.size(); i++) {
        a.push_back(b[i]);
      }
    }
  }
  if (vis[u] > a.size()) {
    f = 1;
  } else {
    a.insert(a.begin() + vis[u], u);
  }
  return a;
}
int main() {
  int n, root;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a) {
      v[i].push_back(node{a, b});
      v[a].push_back(node{i, b});
    } else {
      root = i;
    }
    vis[i] = b;
  }
  vector<int> a = dfs(root, -1);
  if (f) {
    printf("NO\n");
    return 0;
  }
  int p = 1;
  for (int i = 0; i < a.size(); i++) {
    pos[a[i]] = p++;
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%d ", pos[i]);
  }
  return 0;
}
