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;
vector<int> adj[2001];
int n;
int p[2001];
int c[2001];
int a[2001];
vector<int> values[2001];
int subct[2001];
int mx = 1;
void dfs1(int node) {
  subct[node] = 0;
  for (auto it : adj[node]) {
    dfs1(it);
    subct[node] += (1 + subct[it]);
  }
}
void adfs(int node, int val) {
  if (a[node] >= val) a[node]++;
  for (auto &it : values[node]) {
    if (it >= val) it++;
  }
  for (auto it : adj[node]) adfs(it, val);
}
void dfs(int node) {
  if (adj[node].size() == 0) {
    a[node] = mx;
    mx++;
    values[node].push_back(a[node]);
    return;
  }
  for (auto it : adj[node]) {
    dfs(it);
    for (auto it2 : values[it]) values[node].push_back(it2);
  }
  if (c[node] == values[node].size()) {
    values[node].push_back(mx);
    a[node] = mx;
    mx++;
    return;
  }
  int v = values[node][c[node]];
  a[node] = v;
  for (auto it : adj[node]) adfs(it, v);
  values[node].push_back(values[node].back() + 1);
  mx = values[node].back() + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int root;
  for (int i = int(1); i <= int(n); i++) {
    cin >> p[i] >> c[i];
    if (p[i])
      adj[p[i]].push_back(i);
    else
      root = i;
  }
  dfs1(root);
  for (int i = int(1); i <= int(n); i++) {
    if (c[i] > subct[i]) {
      cout << "NO";
      return 0;
    }
  }
  dfs(root);
  cout << "YES\n";
  for (int i = int(1); i <= int(n); i++) cout << a[i] << " ";
}
