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 = 200010;
int n, root;
int vis[N], ans[N], c[N], sz[N];
int h[N], e[N], ne[N], idx;
void add(int a, int b) {
  e[idx] = b;
  ne[idx] = h[a];
  h[a] = idx++;
}
bool dfs(int u) {
  sz[u] = 1;
  int res = 0, y = 0;
  while (1) {
    if (!vis[y + 1]) {
      y++;
      if (res == c[u]) {
        ans[u] = y;
        vis[y] = 1;
        break;
      }
      res++;
    } else
      y++;
  }
  for (int i = h[u]; i != -1; i = ne[i]) {
    dfs(e[i]);
    sz[u] += sz[e[i]];
  }
  if (sz[u] <= c[u]) {
    cout << "NO" << endl;
    exit(0);
  }
  return true;
}
int main() {
  memset(h, -1, sizeof h);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x >> c[i];
    if (x == 0) root = i;
    add(x, i);
  }
  bool suc = dfs(root);
  if (suc) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  } else
    cout << "NO" << endl;
  return 0;
}
