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_MAX = 2002;
int n;
int p[N_MAX], c[N_MAX];
int root;
vector<int> sons[N_MAX];
int sub[N_MAX];
void dfs(int node) {
  sub[node] = 1;
  for (int u : sons[node]) {
    dfs(u);
    sub[node] += sub[u];
  }
}
void f(int node) {
  while (p[node] != 0) {
    node = p[node];
    sub[node]--;
  }
}
int a[N_MAX];
bool solve(int node, int val) {
  if (a[node] == 0 && c[node] == sub[node] - 1) {
    a[node] = val;
    f(node);
    return true;
  }
  for (int u : sons[node])
    if (solve(u, val) == true) return true;
  return false;
}
int check(int node, int th) {
  int ans = 0;
  if (a[node] < th) ans++;
  for (int u : sons[node]) ans += check(u, th);
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    if (p[i] == 0)
      root = i;
    else
      sons[p[i]].push_back(i);
  }
  dfs(root);
  int cnt = n;
  while (cnt > 0) {
    solve(root, cnt);
    cnt--;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] == 0) {
      cout << "NO\n";
      return 0;
    }
  for (int i = 1; i <= n; i++)
    if (check(i, a[i]) != c[i]) {
      cout << "NO\n";
      return 0;
    }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << a[i] << " ";
  cout << "\n";
  return 0;
}
