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;
const int maxn = 2005, maxm = 100005, INF = 0x3f3f3f3f;
inline int read() {
  int out = 0, flag = 1;
  char c = getchar();
  while (c < 48 || c > 57) {
    if (c == '-') flag = 0;
    c = getchar();
  }
  while (c >= 48 && c <= 57) {
    out = (out << 1) + (out << 3) + c - 48;
    c = getchar();
  }
  return flag ? out : -out;
}
int ls[maxn], rb[maxn], fa[maxn], c[maxn], n;
int used[maxn], rt, num[maxn];
void dfs(int u) {
  for (int i = 1, cnt = 0; i <= n; i++) {
    if (!used[i]) cnt++;
    if (cnt == c[u] + 1) {
      num[u] = i;
      used[i] = true;
      break;
    }
  }
  for (int k = ls[u]; k; k = rb[k]) {
    dfs(k);
  }
}
int flag, cal[maxn];
void dfs3(int u) {
  cal[num[u]]++;
  for (int k = ls[u]; k; k = rb[k]) dfs3(k);
}
void dfs2(int u) {
  for (int i = 1; i <= (n); i++) cal[i] = 0;
  for (int k = ls[u]; k; k = rb[k]) dfs3(k);
  int cnt = 0;
  for (int i = 1; i < num[u]; i++) cnt += cal[i];
  if (cnt != c[u]) flag = true;
  for (int k = ls[u]; k; k = rb[k]) dfs2(k);
}
int main() {
  n = read();
  for (int i = 1; i <= (n); i++) {
    fa[i] = read(), c[i] = read();
    rb[i] = ls[fa[i]];
    ls[fa[i]] = i;
    if (!fa[i]) rt = i;
  }
  dfs(rt);
  dfs2(rt);
  if (flag)
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%d ", num[i]);
  }
  return 0;
}
