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;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 2005;
int en, h[N], ans[N], n, ned[N], rt;
vector<int> q;
struct edge {
  int n, v;
} e[N];
void add(int x, int y) {
  e[++en] = (edge){h[x], y};
  h[x] = en;
}
int dfs(int x) {
  int sz = 1;
  for (int i = h[x]; i; i = e[i].n) {
    int y = e[i].v;
    sz += dfs(y);
  }
  if (sz <= ned[x]) {
    puts("NO");
    exit(0);
  }
  q.insert(q.begin() + ned[x], x);
  return sz;
}
signed main() {
  read(n);
  for (int i = 1, x; i <= n; i++) {
    read(x);
    read(ned[i]);
    if (!x)
      rt = i;
    else
      add(x, i);
  }
  dfs(rt);
  puts("YES");
  for (int i = 0; i < q.size(); i++) ans[q[i]] = i + 1;
  for (int i = 1; i <= n; i++) write(ans[i]), putchar(' ');
}
