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
#include <bits/stdc++.h>
using namespace std;
const int N = 2345;
int n, p[N], c[N], ans[N];
vector<int> adj[N];
vector<pair<int, int>> values[N];
inline void dfs(int u) {
  for (int v : adj[u]) {
    dfs(v);
    for (pair<int, int> s : values[v]) {
      values[u].push_back(s);
    }
  }
  if (c[u] > (int)values[u].size()) {
    cout << "NO" << '\n';
    exit(0);
  } else {
    sort(values[u].begin(), values[u].end());
    int val = 0;
    if (c[u] == 0) {
      val = 1;
    } else {
      val = values[u][c[u] - 1].first + 1;
    }
    for (int i = c[u]; i < (int)values[u].size(); i++) {
      values[u][i].first += 2;
    }
    values[u].push_back(make_pair(val, u));
    sort(values[u].begin(), values[u].end());
  }
}
int main() {
  cin >> n;
  int root = -1;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    if (p[i] == 0) {
      root = i;
    }
    adj[p[i]].push_back(i);
  }
  dfs(root);
  for (auto [label, index] : values[root]) {
    ans[index] = label;
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
}
