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 = 2000 + 5;
std::vector<int> G[MAXN];
std::vector<int> ans[MAXN];
int n, rt, lim[MAXN], sz[MAXN];
inline void dfs(int v, int fa = 0) {
  sz[v] = 1;
  for (auto x : G[v]) {
    if (x == fa) continue;
    dfs(x, v);
    sz[v] += sz[x];
    for (auto y : ans[x]) ans[v].push_back(y);
  }
  if (lim[v] > sz[v] - 1) puts("NO"), exit(0);
  ans[v].insert(ans[v].begin() + lim[v], v);
}
int val[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int f;
    scanf("%d%d", &f, lim + i);
    if (!f)
      rt = i;
    else
      G[f].push_back(i), G[i].push_back(f);
  }
  dfs(rt);
  int tot = 0;
  for (auto x : ans[rt]) val[x] = ++tot;
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d ", val[i]);
  puts("");
  return 0;
}
