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<vector<int> > g;
vector<int> v, c, ans, subtree;
int k;
void dfs2(int u, int p) {
  subtree[u] = 1;
  for (int to : g[u]) {
    if (to == p) {
      continue;
    }
    dfs2(to, u);
    subtree[u] += subtree[to];
  }
}
void dfs(int u, int p) {
  ans[u] = v[c[u]];
  v.erase(v.begin() + c[u]);
  for (int to : g[u]) {
    if (to == p) {
      continue;
    }
    dfs(to, u);
  }
}
int main() {
  k = 1;
  int n;
  cin >> n;
  c = vector<int>(n + 2);
  ans = vector<int>(n + 2);
  subtree = vector<int>(n + 2);
  g = vector<vector<int> >(n + 2);
  int root;
  for (int i = 1; i <= n; i++) {
    v.push_back(i);
    int p;
    cin >> p >> c[i];
    if (p == 0) {
      root = i;
    } else {
      g[i].push_back(p);
      g[p].push_back(i);
    }
  }
  dfs2(root, 0);
  for (int i = 1; i <= n; i++) {
    if (c[i] >= subtree[i]) {
      k = 0;
    }
  }
  if (k == 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  dfs(root, 0);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}
