#include <bits/stdc++.h>
using namespace std;
int n, c[200005], p[200005], x, ans[200005], f;
vector<int> v[200005], v1;
void solve(int a, int l) {
  for (int i = 0; i < v[a].size(); i++) {
    if (v[a][i] != l) {
      solve(v[a][i], a);
      p[a] += p[v[a][i]];
    }
  }
  if (p[a] < c[a]) {
    x = 1;
  }
  p[a]++;
}
void find(int a, int l) {
  ans[a] = v1[c[a]];
  v1.erase(v1.begin() + c[a]);
  for (int i = 0; i < v[a].size(); i++) {
    if (v[a][i] != l) {
      find(v[a][i], a);
    }
  }
}
int a;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a >> c[i];
    if (a != 0) {
      v[i].push_back(a);
      v[a].push_back(i);
    } else
      f = i;
    v1.push_back(i);
  }
  solve(f, 0);
  if (x == 1) {
    cout << "NO" << endl;
    return 0;
  }
  find(f, 0);
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}

### Prompt

Your task is to create 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>
using namespace std;
int n, c[200005], p[200005], x, ans[200005], f;
vector<int> v[200005], v1;
void solve(int a, int l) {
  for (int i = 0; i < v[a].size(); i++) {
    if (v[a][i] != l) {
      solve(v[a][i], a);
      p[a] += p[v[a][i]];
    }
  }
  if (p[a] < c[a]) {
    x = 1;
  }
  p[a]++;
}
void find(int a, int l) {
  ans[a] = v1[c[a]];
  v1.erase(v1.begin() + c[a]);
  for (int i = 0; i < v[a].size(); i++) {
    if (v[a][i] != l) {
      find(v[a][i], a);
    }
  }
}
int a;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a >> c[i];
    if (a != 0) {
      v[i].push_back(a);
      v[a].push_back(i);
    } else
      f = i;
    v1.push_back(i);
  }
  solve(f, 0);
  if (x == 1) {
    cout << "NO" << endl;
    return 0;
  }
  find(f, 0);
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}

```