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 N = 1e5 + 10;
const int inf = 1e6;
const int INF = 1e9;
int a[N], b[N];
vector<int> p[N], d[N];
void dfs(int x) {
  for (auto& i : p[x]) {
    dfs(i);
    for (auto& j : d[i]) {
      if ((int)d[x].size() == b[x]) d[x].push_back(x), a[x] = d[x].size();
      d[x].push_back(j);
      a[j] = d[x].size();
    }
  }
  if ((int)d[x].size() == b[x]) d[x].push_back(x), a[x] = d[x].size();
  if ((int)d[x].size() < b[x]) {
    printf("NO\n");
    exit(0);
  }
}
int main() {
  int n, rt;
  scanf("%d", &n);
  for (auto i = (1); i <= (n); ++i) {
    int x;
    scanf("%d%d", &x, &b[i]);
    if (x)
      p[x].push_back(i);
    else
      rt = i;
  }
  dfs(rt);
  printf("YES\n");
  for (auto i = (1); i <= (n); ++i) printf("%d ", a[i]);
}
