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 void read(T &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c & 15), c = getchar();
}
int n, head[2001], cnt, c[2001], a[2001], sz[2001], rt;
bool bb;
struct Edge {
  int a, b;
} e[2001];
inline void add(int a, int b) {
  e[++cnt].a = head[a], e[cnt].b = b, head[a] = cnt;
}
inline void dfs1(int x, int y) {
  if (a[x] >= y) ++a[x];
  for (int i = head[x]; i; i = e[i].a) dfs1(e[i].b, y);
}
inline void dfs2(int x, int y) {
  a[x] += y;
  for (int i = head[x]; i; i = e[i].a) dfs2(e[i].b, y);
}
inline void dfs(int x) {
  for (int i = head[x]; i; i = e[i].a)
    dfs(e[i].b), dfs2(e[i].b, sz[x]), sz[x] += sz[e[i].b];
  ++sz[x];
  if (c[x] >= sz[x]) {
    bb = 1;
    return;
  }
  dfs1(x, c[x] + 1);
  a[x] = c[x] + 1;
}
signed main() {
  read(n);
  for (int i = 1, x; i <= n; i++) {
    read(x), read(c[i]);
    if (x)
      add(x, i);
    else
      rt = i;
  }
  dfs(rt);
  if (bb) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}
