#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long mod = 998244353;
const int phash = 3;
const long long modhash = 1000000000000000003;
vector<int> g[2005];
int cnt[2005];
vector<int> dfs(int v) {
  vector<int> ans;
  for (int i : g[v]) {
    vector<int> temp = dfs(i);
    ans.insert(ans.end(), temp.begin(), temp.end());
  }
  if (ans.size() < cnt[v]) {
    cout << "NO";
    exit(0);
  }
  ans.insert(ans.begin() + cnt[v], v);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int root = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x >> cnt[i];
    if (x == 0)
      root = i;
    else {
      x--;
      g[x].push_back(i);
    }
  }
  vector<int> pos = dfs(root);
  cout << "YES" << '\n';
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[pos[i]] = i + 1;
  for (int i : ans) cout << i << " ";
}

### Prompt

Please formulate a CPP solution 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long mod = 998244353;
const int phash = 3;
const long long modhash = 1000000000000000003;
vector<int> g[2005];
int cnt[2005];
vector<int> dfs(int v) {
  vector<int> ans;
  for (int i : g[v]) {
    vector<int> temp = dfs(i);
    ans.insert(ans.end(), temp.begin(), temp.end());
  }
  if (ans.size() < cnt[v]) {
    cout << "NO";
    exit(0);
  }
  ans.insert(ans.begin() + cnt[v], v);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int root = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x >> cnt[i];
    if (x == 0)
      root = i;
    else {
      x--;
      g[x].push_back(i);
    }
  }
  vector<int> pos = dfs(root);
  cout << "YES" << '\n';
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[pos[i]] = i + 1;
  for (int i : ans) cout << i << " ";
}

```