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 max_n = 2022, inf = 1000111222;
int n, c[max_n], sz[max_n], a[max_n];
vector<int> g[max_n];
void add(int v, int x) {
  a[v] += x;
  for (int to : g[v]) {
    add(to, x);
  }
}
void increase(int v, int x) {
  if (a[v] > x) {
    ++a[v];
  }
  for (int to : g[v]) {
    increase(to, x);
  }
}
void dfs(int v) {
  for (int to : g[v]) {
    dfs(to);
    sz[v] += sz[to];
  }
  if (c[v] > sz[v]) {
    puts("NO");
    exit(0);
  }
  ++sz[v];
  int x = 0;
  for (int to : g[v]) {
    add(to, x);
    x += sz[to];
  }
  for (int to : g[v]) {
    increase(to, c[v]);
  }
  a[v] = c[v] + 1;
}
int main() {
  int root;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int p;
    scanf("%d%d", &p, &c[i]);
    --p;
    if (p == -1) {
      root = i;
    } else {
      g[p].push_back(i);
    }
  }
  dfs(root);
  puts("YES");
  for (int i = 0; i < n; ++i) {
    printf("%d ", a[i]);
  }
  puts("");
  return 0;
}
