#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3 + 5;
vector<int> edg[maxn];
int vis[maxn], res[maxn];
int sze[maxn];
int c[maxn];
int n;
void dfs_sz(int x) {
  sze[x] = 1;
  int len = edg[x].size();
  for (int i = 0; i < len; i++) {
    dfs_sz(edg[x][i]);
    sze[x] += sze[edg[x][i]];
  }
}
void dfs(int x) {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) cnt++;
    if (cnt == c[x] + 1) {
      res[x] = i;
      break;
    }
  }
  vis[res[x]] = 1;
  int len = edg[x].size();
  for (int i = 0; i < len; i++) {
    dfs(edg[x][i]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int rt;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x >> c[i];
    if (x == 0)
      rt = i;
    else
      edg[x].push_back(i);
  }
  dfs_sz(rt);
  for (int i = 1; i <= n; i++) {
    if (sze[i] <= c[i]) {
      cout << "NO\n";
      return 0;
    }
  }
  dfs(rt);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << res[i] << ' ';
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 2e3 + 5;
vector<int> edg[maxn];
int vis[maxn], res[maxn];
int sze[maxn];
int c[maxn];
int n;
void dfs_sz(int x) {
  sze[x] = 1;
  int len = edg[x].size();
  for (int i = 0; i < len; i++) {
    dfs_sz(edg[x][i]);
    sze[x] += sze[edg[x][i]];
  }
}
void dfs(int x) {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) cnt++;
    if (cnt == c[x] + 1) {
      res[x] = i;
      break;
    }
  }
  vis[res[x]] = 1;
  int len = edg[x].size();
  for (int i = 0; i < len; i++) {
    dfs(edg[x][i]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int rt;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x >> c[i];
    if (x == 0)
      rt = i;
    else
      edg[x].push_back(i);
  }
  dfs_sz(rt);
  for (int i = 1; i <= n; i++) {
    if (sze[i] <= c[i]) {
      cout << "NO\n";
      return 0;
    }
  }
  dfs(rt);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << res[i] << ' ';
  }
}

```