#include <bits/stdc++.h>
using namespace std;
const long long N = 2005;
vector<long long> g[N];
long long tin[N], tout[N], tt = 1;
long long c[N], res[N], n, tt2 = 1;
bool ok = 1;
void dfs(long long u, long long p) {
  tin[u] = tt++;
  for (auto v : g[u])
    if (v != p) dfs(v, u);
  tout[u] = tt++;
  vector<pair<long long, long long> > sub;
  for (long long i = 1; i <= n; i++)
    if (u != i && tin[u] <= tin[i] && tout[i] <= tout[u])
      sub.push_back({res[i], i});
  sort(sub.begin(), sub.end());
  if (sub.size() < c[u])
    ok = 0;
  else if (sub.size() == c[u])
    res[u] = tt2;
  else {
    res[u] = sub[c[u]].first;
    for (long long i = c[u]; i < sub.size(); i++) res[sub[i].second]++;
  }
  tt2++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long p;
    cin >> p >> c[i];
    g[p].push_back(i);
  }
  long long r = g[0][0];
  dfs(r, 0);
  if (!ok)
    cout << "NO";
  else {
    cout << "YES\n";
    for (long long i = 1; i <= n; i++) cout << res[i] << " ";
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 long long N = 2005;
vector<long long> g[N];
long long tin[N], tout[N], tt = 1;
long long c[N], res[N], n, tt2 = 1;
bool ok = 1;
void dfs(long long u, long long p) {
  tin[u] = tt++;
  for (auto v : g[u])
    if (v != p) dfs(v, u);
  tout[u] = tt++;
  vector<pair<long long, long long> > sub;
  for (long long i = 1; i <= n; i++)
    if (u != i && tin[u] <= tin[i] && tout[i] <= tout[u])
      sub.push_back({res[i], i});
  sort(sub.begin(), sub.end());
  if (sub.size() < c[u])
    ok = 0;
  else if (sub.size() == c[u])
    res[u] = tt2;
  else {
    res[u] = sub[c[u]].first;
    for (long long i = c[u]; i < sub.size(); i++) res[sub[i].second]++;
  }
  tt2++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long p;
    cin >> p >> c[i];
    g[p].push_back(i);
  }
  long long r = g[0][0];
  dfs(r, 0);
  if (!ok)
    cout << "NO";
  else {
    cout << "YES\n";
    for (long long i = 1; i <= n; i++) cout << res[i] << " ";
  }
  return 0;
}

```