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 = 2100;
int n, fa[N], s[N], root, size[N], ans[N];
vector<int> to[N], a[N];
template <class T>
inline void read(T &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
  return;
}
void dfs1(int now) {
  size[now] = 1;
  if (!s[now]) a[now].push_back(now);
  for (auto &v : to[now]) {
    dfs1(v), size[now] += size[v];
    for (auto &k : a[v]) {
      a[now].push_back(k);
      if (a[now].size() == s[now]) a[now].push_back(now);
    }
  }
  if (size[now] <= s[now]) {
    printf("NO\n");
    exit(0);
  }
  return;
}
int main() {
  read(n);
  for (register int i = 1; i <= n; ++i) {
    read(fa[i]), read(s[i]);
    if (fa[i])
      to[fa[i]].push_back(i);
    else
      root = i;
  }
  dfs1(root);
  printf("YES\n");
  for (register int i = 0; i < n; ++i) ans[a[root][i]] = i + 1;
  for (register int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
