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>
namespace local {
const int kN = 2005;
std::vector<int> son[kN];
int n;
int c[kN], size[kN], ans[kN];
bool used[kN], flag = false;
void get_size(int u) {
  size[u] = 1;
  for (std::vector<int>::iterator i = son[u].begin(); i != son[u].end(); i++) {
    get_size(*i);
    size[u] += size[*i];
  }
}
void dfs(int u) {
  if (c[u] >= size[u]) {
    flag = true;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      cnt++;
    }
    if (cnt == c[u] + 1) {
      used[i] = true;
      ans[u] = i;
      break;
    }
  }
  for (std::vector<int>::iterator i = son[u].begin(); i != son[u].end(); i++) {
    dfs(*i);
  }
}
}  // namespace local
int main() {
  using namespace local;
  int root = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d%d", &p, c + i);
    if (p) {
      son[p].push_back(i);
    } else {
      root = i;
    }
  }
  get_size(root);
  dfs(root);
  if (flag) {
    puts("NO");
  } else {
    puts("YES");
    for (int i = 1; i <= n; i++) {
      printf("%d ", ans[i]);
    }
    putchar('\n');
  }
  return 0;
}
