#include <bits/stdc++.h>
using namespace std;
const int N = 2000;
int n, c[N], ans[N], sz[N];
vector<int> g[N];
void zhfs(int v) {
  sz[v] = 1;
  for (auto u : g[v]) {
    zhfs(u);
    sz[v] += sz[u];
  }
  if (sz[v] <= c[v]) {
    cout << "NO\n";
    exit(0);
  }
}
void dfs(int v, vector<int> a) {
  ans[v] = a[c[v]];
  a.erase(a.begin() + c[v], a.begin() + c[v] + 1);
  int pos = 0;
  for (auto u : g[v]) {
    vector<int> b;
    while (b.size() < sz[u]) b.push_back(a[pos++]);
    dfs(u, b);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int root = -1;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i];
    p--;
    if (p != -1) {
      g[p].push_back(i);
    } else
      root = i;
  }
  zhfs(root);
  cout << "YES\n";
  vector<int> a(n);
  iota((a).begin(), (a).end(), 1);
  dfs(root, a);
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

### Prompt

Please formulate 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 N = 2000;
int n, c[N], ans[N], sz[N];
vector<int> g[N];
void zhfs(int v) {
  sz[v] = 1;
  for (auto u : g[v]) {
    zhfs(u);
    sz[v] += sz[u];
  }
  if (sz[v] <= c[v]) {
    cout << "NO\n";
    exit(0);
  }
}
void dfs(int v, vector<int> a) {
  ans[v] = a[c[v]];
  a.erase(a.begin() + c[v], a.begin() + c[v] + 1);
  int pos = 0;
  for (auto u : g[v]) {
    vector<int> b;
    while (b.size() < sz[u]) b.push_back(a[pos++]);
    dfs(u, b);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int root = -1;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i];
    p--;
    if (p != -1) {
      g[p].push_back(i);
    } else
      root = i;
  }
  zhfs(root);
  cout << "YES\n";
  vector<int> a(n);
  iota((a).begin(), (a).end(), 1);
  dfs(root, a);
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

```