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 long long N = 5e3 + 10, mod = 1e9 + 7, inf = 1e18;
int p[N], c[N], root, n, check = 1, ans[N];
vector<int> G[N];
vector<pair<int, int> > vt[N];
void dfs(int u) {
  for (int v : G[u]) {
    dfs(v);
    int tmp;
    if (vt[u].empty())
      tmp = 0;
    else
      tmp = vt[u].back().first;
    for (pair<int, int> x : vt[v])
      vt[u].push_back(make_pair(x.first + tmp, x.second));
  }
  if (c[u] > (int)vt[u].size()) {
    check = 0;
    return;
  }
  vector<pair<int, int> > tmp;
  for (int i = 0; i < c[u]; i++) tmp.push_back(vt[u][i]);
  tmp.push_back(make_pair(c[u] + 1, u));
  for (int i = c[u]; i < (int)vt[u].size(); i++)
    tmp.push_back(make_pair(vt[u][i].first + 1, vt[u][i].second));
  vt[u] = tmp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    if (!p[i]) root = i;
    G[p[i]].push_back(i);
  }
  dfs(root);
  if (!check) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (pair<int, int> x : vt[root]) ans[x.second] = x.first;
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
