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 MAX = 2e3 + 3;
vector<int> G[MAX], rk[MAX];
int rt, n;
int c[MAX], a[MAX];
bool dfs(int x) {
  vector<int> srk;
  for (int i = 0; i < G[x].size(); i++) {
    int son = G[x][i];
    if (!dfs(son)) return 0;
    for (int j = 0; j < rk[son].size(); j++) srk.push_back(rk[son][j]);
  }
  if (c[x] > srk.size()) return 0;
  for (int i = 0; i < c[x]; i++) rk[x].push_back(srk[i]);
  rk[x].push_back(x);
  for (int i = c[x]; i < srk.size(); i++) rk[x].push_back(srk[i]);
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d%d", &p, &c[i]);
    if (!p) rt = i;
    G[p].push_back(i);
  }
  if (dfs(rt)) {
    printf("YES\n");
    for (int i = 0; i < n; i++) a[rk[rt][i]] = i + 1;
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  } else
    printf("NO\n");
}
