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;
using ll = long long;
int n, arr[2001], a, b, num[2001];
vector<int> adj[2001];
bool good = true;
vector<int> dfs(int src = b, int p = 0) {
  int cnt = 0;
  vector<int> chs;
  for (int ch : adj[src]) {
    if (ch == p) {
      continue;
    }
    for (int x : dfs(ch, src)) {
      if (cnt == arr[src]) {
        chs.emplace_back(src);
        cnt++;
      }
      chs.emplace_back(x);
      cnt++;
    }
  }
  if (arr[src] > cnt) {
    good = false;
  }
  if (chs.empty()) {
    if (arr[src]) {
      good = false;
    }
    chs.emplace_back(src);
  } else if (cnt == arr[src]) {
    chs.emplace_back(src);
  }
  return chs;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &a, &arr[i]);
    if (a) {
      adj[a].emplace_back(i);
      adj[i].emplace_back(a);
    } else {
      b = i;
    }
  }
  vector<int> ans = dfs();
  if (!good) {
    puts("NO");
    return 0;
  }
  int c = 1;
  for (int x : ans) {
    num[x] = c++;
  }
  puts("YES");
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", num[i], " \n"[i == n]);
  }
  return 0;
}
