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>
const int maxn = 2e3 + 3;
int n, c[maxn], p[maxn], a[maxn], s[maxn];
std::vector<int> E[maxn], U;
void dfsA(int u) {
  s[u] = 1;
  for (auto &&v : E[u]) {
    dfsA(v);
    s[u] += s[v];
  }
}
void dfsB(int u) {
  if (c[u] >= s[u]) {
    puts("NO");
    exit(0);
  }
  a[u] = U[c[u]];
  U.erase(U.begin() + c[u]);
  for (auto &&v : E[u]) {
    dfsB(v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", p + i, c + i);
    E[p[i]].push_back(i);
  }
  for (int i = 0; i <= n; ++i) {
    U.push_back(i);
  }
  dfsA(0);
  dfsB(0);
  puts("YES");
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", a[i], " \n"[i == n]);
  }
  return 0;
}
