#include <bits/stdc++.h>
using namespace std;
int num[2030], a[2030];
set<int> se;
int n, s, sc[2030];
vector<int> v[2030];
void dfs(int node) {
  num[node]++;
  for (auto pp : v[node]) {
    dfs(pp);
    num[node] += num[pp];
  }
}
void solve(int node) {
  auto it = se.begin();
  while (a[node]--) it++;
  sc[node] = *it;
  se.erase(it);
  for (auto pp : v[node]) solve(pp);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d%d", &p, &a[i]);
    if (p)
      v[p].push_back(i);
    else
      s = i;
  }
  dfs(s);
  for (int i = 1; i <= n; i++) {
    if (a[i] >= num[i]) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) se.insert(i);
  solve(s);
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", sc[i]);
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int num[2030], a[2030];
set<int> se;
int n, s, sc[2030];
vector<int> v[2030];
void dfs(int node) {
  num[node]++;
  for (auto pp : v[node]) {
    dfs(pp);
    num[node] += num[pp];
  }
}
void solve(int node) {
  auto it = se.begin();
  while (a[node]--) it++;
  sc[node] = *it;
  se.erase(it);
  for (auto pp : v[node]) solve(pp);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d%d", &p, &a[i]);
    if (p)
      v[p].push_back(i);
    else
      s = i;
  }
  dfs(s);
  for (int i = 1; i <= n; i++) {
    if (a[i] >= num[i]) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) se.insert(i);
  solve(s);
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", sc[i]);
  return 0;
}

```