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;
int n, fa[2005], c[2005], siz[2005], id[2005], dfn[2005], tot;
vector<int> G[2005];
int vis[2005];
int tmp[2005], cnt = 0;
void dfs(int x) {
  siz[x] = 1;
  dfn[x] = ++tot;
  for (int i = 0; i < G[x].size(); i++) {
    int V = G[x][i];
    dfs(V);
    siz[x] += siz[V];
  }
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (dfn[x] < dfn[i] && dfn[i] <= dfn[x] + siz[x] - 1) tmp[++cnt] = id[i];
  }
  sort(tmp + 1, tmp + cnt + 1);
  id[x] = tmp[c[x]] + 1;
  for (int i = 1; i <= n; i++) {
    if (i == x) continue;
    if (id[i] >= id[x]) id[i]++;
  }
}
int root;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &fa[i], &c[i]);
    if (fa[i] == 0)
      root = i;
    else
      G[fa[i]].push_back(i);
  }
  dfs(root);
  for (int i = 1; i <= n; i++) {
    if (c[i] >= siz[i]) {
      printf("NO");
      return 0;
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%d ", id[i]);
  }
  return 0;
}
