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;
using int64 = long long;
void setup_io() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
}
const int N = 3e5 + 10;
int n;
vector<int> adj[N];
int c[N];
int v[N];
int id = 1;
bool vis[N];
bool dfs(int cur) {
  if (!vis[cur] && c[cur] == 0) {
    vis[cur] = 1;
    v[cur] = id++;
    return true;
  }
  for (const auto i : adj[cur]) {
    if (dfs(i)) {
      if (!vis[cur]) {
        --c[cur];
      }
      return true;
    }
  }
  return false;
}
bool check() {
  memset(vis, 0, sizeof(vis));
  int root = adj[0][0];
  for (int i = 0; i < n; ++i) {
    if (!dfs(root)) return false;
  }
  return true;
}
int main() {
  setup_io();
  cin >> n;
  int p;
  for (int i = 1; i <= n; ++i) {
    cin >> p >> c[i];
    adj[p].push_back(i);
  }
  if (check()) {
    cout << "YES\n";
    for (int i = 1; i <= n; ++i) {
      cout << v[i] << " ";
    }
    cout << "\n";
  } else {
    cout << "NO\n";
  }
}
