#include <bits/stdc++.h>
using namespace std;
const int maxn = 2005;
int n;
vector<int> graph[maxn];
int c[maxn];
void fail() {
  cout << "NO\n";
  exit(0);
}
vector<int> dfs(int v) {
  vector<int> ans;
  for (auto to : graph[v]) {
    for (auto it : dfs(to)) {
      if (((int)(ans).size()) == c[v]) {
        ans.push_back(v);
      }
      ans.push_back(it);
    }
  }
  if (((int)(ans).size()) == c[v]) {
    ans.push_back(v);
  }
  if (((int)(ans).size()) < c[v]) {
    fail();
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int root = -1;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p >> c[i];
    if (p == 0) {
      root = i;
    } else {
      graph[p].push_back(i);
    }
  }
  vector<int> vec = dfs(root);
  vector<int> ans(n + 1);
  for (int i = 0; i < n; ++i) {
    ans[vec[i]] = i + 1;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}

### 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;
const int maxn = 2005;
int n;
vector<int> graph[maxn];
int c[maxn];
void fail() {
  cout << "NO\n";
  exit(0);
}
vector<int> dfs(int v) {
  vector<int> ans;
  for (auto to : graph[v]) {
    for (auto it : dfs(to)) {
      if (((int)(ans).size()) == c[v]) {
        ans.push_back(v);
      }
      ans.push_back(it);
    }
  }
  if (((int)(ans).size()) == c[v]) {
    ans.push_back(v);
  }
  if (((int)(ans).size()) < c[v]) {
    fail();
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int root = -1;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p >> c[i];
    if (p == 0) {
      root = i;
    } else {
      graph[p].push_back(i);
    }
  }
  vector<int> vec = dfs(root);
  vector<int> ans(n + 1);
  for (int i = 0; i < n; ++i) {
    ans[vec[i]] = i + 1;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}

```