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 maxn = 1e5;
struct node {
  int id;
  int sz;
  int cd;
  vector<int> s;
};
node nd[maxn];
vector<int> G[maxn];
int root = -1;
bool flag = true;
void dfs1(int v, int f) {
  nd[v].sz += 1;
  for (int i = 0; i < G[v].size(); i++) {
    int to = G[v][i];
    if (to == f) continue;
    dfs1(to, v);
    nd[v].sz += nd[to].sz;
  }
  if (nd[v].cd > nd[v].sz - 1) flag = false;
}
int ans[maxn];
void dfs2(int v, int f) {
  ans[v] = nd[v].s[nd[v].cd];
  nd[v].s.erase(nd[v].s.begin() + nd[v].cd);
  int pos = 0;
  for (int i = 0; i < G[v].size(); i++) {
    int to = G[v][i];
    if (to == f) continue;
    for (int i = pos; i < pos + nd[to].sz; i++) nd[to].s.push_back(nd[v].s[i]);
    pos += nd[to].sz;
    dfs2(to, v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    int x;
    int cd;
    cin >> x >> cd;
    if (x == 0) root = i;
    G[x].push_back(i);
    G[i].push_back(x);
    nd[i].cd = cd;
    nd[i].id = i;
  }
  dfs1(root, 0);
  if (flag == false) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= N; i++) nd[root].s.push_back(i);
  dfs2(root, 0);
  for (int i = 1; i <= N; i++) cout << ans[i] << ' ';
  return 0;
}
