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, cnt, tot, rt, a[2010], fa[2010], e[2010], hd[2010], nt[2010], ru[2010],
    chu[2010], sz[2010], c[2010], flag;
void build(int x, int y) {
  tot++;
  e[tot] = y;
  nt[tot] = hd[x];
  hd[x] = tot;
}
void dfs(int x) {
  int i, j, la = 0;
  ru[x] = ++cnt;
  sz[x] = 1;
  for (i = hd[x]; i; i = nt[i]) {
    dfs(e[i]);
    sz[x] += sz[e[i]];
  }
  for (i = hd[x]; i; i = nt[i]) {
    for (j = 1; j <= n; j++)
      if (ru[e[i]] <= ru[j] && ru[j] <= chu[e[i]]) a[j] += la;
    la += sz[e[i]];
  }
  chu[x] = cnt;
  if (c[x] >= sz[x]) {
    flag = 1;
    return;
  }
  for (i = 1; i <= n; i++)
    if (ru[x] <= ru[i] && ru[i] <= chu[x] && a[i] > c[x]) a[i]++;
  a[x] = c[x] + 1;
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &fa[i], &c[i]);
    if (!fa[i])
      rt = i;
    else
      build(fa[i], i);
  }
  dfs(rt);
  if (flag)
    puts("NO");
  else {
    puts("YES");
    for (i = 1; i <= n; i++) printf("%d ", a[i]);
  }
}
