#include <bits/stdc++.h>
using namespace std;
const int maxn = 2009;
int n, c[maxn], root, l, temp[maxn], ans[maxn], flag, Size[maxn];
vector<int> vec[maxn];
int find(int x) {
  int t = 0;
  for (int i = 1; i <= n; i++) {
    if (temp[i] == 0) continue;
    t++;
    if (t == x) {
      temp[i] = 0;
      return i;
    }
  }
  return -1;
}
void dfs(int now) {
  Size[now] = 1;
  ans[now] = find(c[now] + 1);
  if (ans[now] == -1) flag = -1;
  for (int i = 0; i < vec[now].size(); i++) {
    int x = vec[now][i];
    dfs(x);
    Size[now] += Size[x];
  }
  if (Size[now] - 1 < c[now]) flag = -1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) temp[i] = i;
  for (int i = 1; i <= n; i++) {
    cin >> l >> c[i];
    vec[l].push_back(i);
    if (l == 0) root = i;
  }
  dfs(root);
  if (flag == -1)
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 2009;
int n, c[maxn], root, l, temp[maxn], ans[maxn], flag, Size[maxn];
vector<int> vec[maxn];
int find(int x) {
  int t = 0;
  for (int i = 1; i <= n; i++) {
    if (temp[i] == 0) continue;
    t++;
    if (t == x) {
      temp[i] = 0;
      return i;
    }
  }
  return -1;
}
void dfs(int now) {
  Size[now] = 1;
  ans[now] = find(c[now] + 1);
  if (ans[now] == -1) flag = -1;
  for (int i = 0; i < vec[now].size(); i++) {
    int x = vec[now][i];
    dfs(x);
    Size[now] += Size[x];
  }
  if (Size[now] - 1 < c[now]) flag = -1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) temp[i] = i;
  for (int i = 1; i <= n; i++) {
    cin >> l >> c[i];
    vec[l].push_back(i);
    if (l == 0) root = i;
  }
  dfs(root);
  if (flag == -1)
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  }
}

```