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 = 4e3 + 10;
const int M = 1e4 + 10;
int n;
int a[N];
int h[N], e[M], ne[M], idx;
int c[N];
void add(int a, int b) {
  e[idx] = b;
  ne[idx] = h[a];
  h[a] = idx++;
}
vector<int> dfs(int u, int fa) {
  vector<int> ans;
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i];
    if (j == fa) continue;
    vector<int> tmp = dfs(j, u);
    for (int i = 0; i < tmp.size(); i++) ans.push_back(tmp[i]);
  }
  if (u == 0) return ans;
  if (ans.size() < c[u]) {
    puts("NO");
    exit(0);
  }
  vector<int> now;
  for (int i = 0; i < c[u]; i++) {
    now.push_back(ans[i]);
  }
  now.push_back(u);
  for (int i = c[u]; i < ans.size(); i++) now.push_back(ans[i]);
  ans = now;
  return ans;
}
int main() {
  memset(h, -1, sizeof(h));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d%d", &x, &c[i]);
    add(x, i);
  }
  vector<int> ans = dfs(0, -1);
  puts("YES");
  for (int i = 0; i < ans.size(); i++) a[ans[i]] = i + 1;
  for (int i = 1; i <= n; i++) printf("%d%c", a[i], i == n ? '\n' : ' ');
}
