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 = 2e3 + 3;
int t[MAXN];
int ans[MAXN];
set<int> q;
vector<int> v[MAXN];
int preDFS(int x) {
  int ret = 0;
  for (auto it : v[x]) ret += preDFS(it);
  if (ret < t[x]) {
    printf("NO");
    exit(0);
  }
  return ret + 1;
}
void DFS(int x) {
  if (t[x] == 0) {
    ans[x] = *q.begin();
    q.erase(*q.begin());
  } else {
    int licz = 0;
    for (auto it : q) {
      if (licz == t[x]) {
        ans[x] = it;
        break;
      }
      licz++;
    }
    q.erase(ans[x]);
  }
  for (auto it : v[x]) DFS(it);
}
int main() {
  int n, p, x = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &p, &t[i]);
    if (p == 0) x = i;
    v[p].push_back(i);
    q.insert(i);
  }
  preDFS(x);
  DFS(x);
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
