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 = (int)2005, mod = (int)0;
int sz[N], c[N], res[N];
vector<int> adj[N];
void dfs_init(int v, int p = -1) {
  sz[v] = 1;
  for (int u : adj[v]) {
    dfs_init(u, v);
    sz[v] += sz[u];
  }
}
void dfs_res(int v, vector<int> all) {
  sort((all).begin(), (all).end());
  res[v] = all[c[v]];
  vector<int> nxt;
  for (int first : all)
    if (first != res[v]) nxt.push_back(first);
  for (int u : adj[v]) {
    vector<int> go;
    while ((int)go.size() != sz[u]) go.push_back(nxt.back()), nxt.pop_back();
    dfs_res(u, go);
  }
}
int main() {
  int n, root = 0;
  cin >> n;
  for (int j = 0; j < n; ++j) {
    int p;
    cin >> p >> c[j];
    --p;
    if (p == -1)
      root = j;
    else
      adj[p].push_back(j);
  }
  vector<int> all;
  for (int j = 0; j < n; ++j) all.push_back(j);
  dfs_init(root);
  for (int v = 0; v < n; ++v)
    if (sz[v] - 1 < c[v]) {
      cout << "NO\n";
      return 0;
    }
  dfs_res(root, all);
  cout << "YES\n";
  for (int j = 0; j < n; ++j) cout << 1 + res[j] << ' ';
}
