#include <bits/stdc++.h>
using namespace std;
const int mx = 2e3 + 9;
void merge(vector<int>& a, vector<int>& b) {
  for (auto x : b) a.push_back(x);
}
int c[mx];
int ans[mx];
vector<int> adj[mx];
vector<int> order[mx];
void dfs(int u, int p = -1) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
    merge(order[u], order[v]);
  }
  if (c[u] > order[u].size()) {
    puts("NO");
    exit(0);
  }
  order[u].insert(order[u].begin() + c[u], u);
}
int main() {
  int n;
  scanf("%d", &n);
  int root = -1;
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d %d", &p, &c[i]);
    if (p) {
      adj[p].push_back(i);
      adj[i].push_back(p);
    } else
      root = i;
  }
  dfs(root);
  puts("YES");
  for (int i = 0; i < n; i++) ans[order[root][i]] = i;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i] + 1);
  puts("");
}

### Prompt

Your challenge is to write 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;
const int mx = 2e3 + 9;
void merge(vector<int>& a, vector<int>& b) {
  for (auto x : b) a.push_back(x);
}
int c[mx];
int ans[mx];
vector<int> adj[mx];
vector<int> order[mx];
void dfs(int u, int p = -1) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
    merge(order[u], order[v]);
  }
  if (c[u] > order[u].size()) {
    puts("NO");
    exit(0);
  }
  order[u].insert(order[u].begin() + c[u], u);
}
int main() {
  int n;
  scanf("%d", &n);
  int root = -1;
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d %d", &p, &c[i]);
    if (p) {
      adj[p].push_back(i);
      adj[i].push_back(p);
    } else
      root = i;
  }
  dfs(root);
  puts("YES");
  for (int i = 0; i < n; i++) ans[order[root][i]] = i;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i] + 1);
  puts("");
}

```