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 = 2005;
vector<int> c(N);
vector<int> sz(N);
vector<int> order[N];
vector<int> g[N];
bool possible = true;
void dfs(int u, int p) {
  sz[u] = 1;
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u);
      order[u].insert(order[u].end(), order[v].begin(), order[v].end());
      sz[u] += sz[v];
    }
  }
  if (c[u] > sz[u] - 1) {
    cout << "NO" << endl;
    exit(0);
  }
  order[u].insert(order[u].begin() + c[u], u);
}
int main() {
  int n;
  cin >> n;
  int root;
  for (int i = 0; i < n; i++) {
    int _p, _c;
    cin >> _p >> _c;
    c[i] = _c;
    if (_p) {
      g[i].push_back(_p - 1);
      g[_p - 1].push_back(i);
    } else {
      root = i;
    }
  }
  dfs(root, root);
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[order[root][i]] = i;
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << ans[i] + 1 << " ";
  cout << endl;
  return 0;
}
