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 N = 2004;
struct E {
  int next, to;
} e[N << 1];
int head[N], c[N], tot = 0, n, m = 0;
void add(int x, int y) {
  e[tot] = E{y, head[x]};
  head[x] = tot++;
}
bool f = 1;
vector<int> dfs(int u) {
  if (!f) return vector<int>{};
  int v;
  vector<int> w, t;
  for (int i = head[u]; ~i; i = e[i].to) {
    t = dfs(e[i].next);
    for (int j = 0; j < t.size(); j++) w.push_back(t[j]);
  }
  if (c[u] > w.size()) {
    f = 0;
    return vector<int>{};
  }
  w.insert(w.begin() + c[u], u);
  return w;
}
int a[N];
int main() {
  memset(head, -1, sizeof(head));
  int x, y, r = -1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &x, &c[i]);
    if (x == 0)
      r = i;
    else
      add(x, i);
  }
  vector<int> v = dfs(r);
  if (!f) return puts("NO"), 0;
  for (int i = 0; i < v.size(); i++) a[v[i]] = i + 1;
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
}
