#include <bits/stdc++.h>
using namespace std;
int inf = 1 << 29;
vector<int> solve(int node, int parent, vector<vector<int> >& tree,
                  vector<int>& r) {
  int sz = 0;
  vector<int> ans;
  for (int neigh : tree[node]) {
    if (neigh == parent) continue;
    vector<int> res = solve(neigh, node, tree, r);
    ans.insert(ans.end(), res.begin(), res.end());
  }
  if (ans.size() < r[node]) {
    cout << "NO" << endl;
    exit(0);
  } else {
    ans.insert(ans.begin() + r[node], node);
    return ans;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> r(n);
  vector<vector<int> > tree(n);
  int root;
  for (int i = 0; i < n; i++) {
    int p, c;
    cin >> p >> c;
    p--;
    r[i] = c;
    if (p != -1) {
      tree[i].push_back(p);
      tree[p].push_back(i);
    } else {
      root = i;
    }
  }
  vector<int> order = solve(root, -1, tree, r);
  vector<int> ans(n);
  for (int i = 0; i < n; i++) {
    ans[order[i]] = i + 1;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
}

### Prompt

Create a solution in CPP for 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 inf = 1 << 29;
vector<int> solve(int node, int parent, vector<vector<int> >& tree,
                  vector<int>& r) {
  int sz = 0;
  vector<int> ans;
  for (int neigh : tree[node]) {
    if (neigh == parent) continue;
    vector<int> res = solve(neigh, node, tree, r);
    ans.insert(ans.end(), res.begin(), res.end());
  }
  if (ans.size() < r[node]) {
    cout << "NO" << endl;
    exit(0);
  } else {
    ans.insert(ans.begin() + r[node], node);
    return ans;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> r(n);
  vector<vector<int> > tree(n);
  int root;
  for (int i = 0; i < n; i++) {
    int p, c;
    cin >> p >> c;
    p--;
    r[i] = c;
    if (p != -1) {
      tree[i].push_back(p);
      tree[p].push_back(i);
    } else {
      root = i;
    }
  }
  vector<int> order = solve(root, -1, tree, r);
  vector<int> ans(n);
  for (int i = 0; i < n; i++) {
    ans[order[i]] = i + 1;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
}

```