#include <bits/stdc++.h>
const long long int MOD = 1e9 + 7;
using namespace std;
int cnt[2005];
std::vector<int> adj[2005];
std::vector<int> dfs(int u) {
  std::vector<int> ret;
  for (int v : adj[u]) {
    std::vector<int> tmp = dfs(v);
    for (int x : tmp) ret.push_back(x);
  }
  if (cnt[u] > ret.size()) {
    cout << "NO" << '\n';
    exit(0);
  }
  ret.insert(ret.begin() + cnt[u], u);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, root;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p, c;
    cin >> p >> c;
    adj[p].push_back(i);
    if (p == 0) root = i;
    cnt[i] = c;
  }
  std::vector<int> v = dfs(root);
  int ans[n + 1];
  for (int i = 1; i <= n; i++) {
    ans[v[i - 1]] = i;
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
const long long int MOD = 1e9 + 7;
using namespace std;
int cnt[2005];
std::vector<int> adj[2005];
std::vector<int> dfs(int u) {
  std::vector<int> ret;
  for (int v : adj[u]) {
    std::vector<int> tmp = dfs(v);
    for (int x : tmp) ret.push_back(x);
  }
  if (cnt[u] > ret.size()) {
    cout << "NO" << '\n';
    exit(0);
  }
  ret.insert(ret.begin() + cnt[u], u);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, root;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p, c;
    cin >> p >> c;
    adj[p].push_back(i);
    if (p == 0) root = i;
    cnt[i] = c;
  }
  std::vector<int> v = dfs(root);
  int ans[n + 1];
  for (int i = 1; i <= n; i++) {
    ans[v[i - 1]] = i;
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}

```