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;
class GRAPH {
 public:
  int to[2010000], next[2010000];
  int head[1001000];
  int cnt;
  GRAPH() { cnt = 1; }
  void add(int x, int y) {
    next[cnt] = head[x];
    to[cnt] = y;
    head[x] = cnt++;
  }
};
GRAPH g;
int pa[1001000];
int c[1001000];
int n, root;
int size[1001000];
int ans[1001000];
vector<int> v[1001000];
void init() {
  int x, y;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    g.add(x, i);
    pa[i] = x;
    c[i] = y;
    if (pa[i] == 0) root = i;
  }
}
void dfs(int pos) {
  size[pos] = 1;
  for (register int i = g.head[pos]; i; i = g.next[i]) {
    dfs(g.to[i]);
    size[pos] += size[g.to[i]];
  }
  if (c[pos] + 1 > size[pos]) {
    cout << "NO";
    exit(0);
  }
}
void dfs1(int pos) {
  for (register int i = g.head[pos]; i; i = g.next[i]) {
    dfs1(g.to[i]);
    for (auto p : v[g.to[i]]) v[pos].push_back(p);
  }
  v[pos].insert(v[pos].begin() + c[pos], pos);
}
int main() {
  init();
  dfs(root);
  dfs1(root);
  int cnt = 1;
  for (auto p : v[root]) ans[p] = cnt++;
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return ~~(0 ^ 0 ^ 0);
}
