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;
int n, a[100005];
void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
void dec(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000007;
}
vector<int> G[100005];
int root, p[100005], c[100005], dep[100005];
bool used[100005];
int tot;
void dfs(int v, int d) {
  dep[v] = d;
  for (auto to : G[v]) dfs(to, d + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p[i], &c[i]);
    if (!p[i])
      root = i;
    else
      G[p[i]].push_back(i);
  }
  memset(used, false, sizeof(used));
  dfs(root, 0);
  for (int i = 1; i <= n; i++) {
    int id = -1, dd = 1000000000;
    for (int j = 1; j <= n; j++) {
      if (!used[j] && c[j] == 0 && dep[j] < dd) {
        id = j;
        dd = dep[j];
      }
    }
    if (id == -1) {
      puts("NO");
      return 0;
    }
    used[id] = true;
    a[id] = ++tot;
    int now = p[id];
    while (now) {
      if (!used[now]) {
        c[now]--;
        if (c[now] < 0) {
          puts("NO");
          return 0;
        }
      }
      now = p[now];
    }
  }
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}
