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 = 2020;
int N;
int a, b;
int root;
vector<int> graf[maxn];
vector<int> manji[maxn];
vector<pair<int, int> > rj[maxn];
int val[maxn];
int M[maxn];
void dfs(int x) {
  if (graf[x].size() == 0) val[x] = 1;
  for (int i = 0; i < graf[x].size(); i++) {
    int n = graf[x][i];
    dfs(n);
    for (int j = 0; j < manji[n].size(); j++) {
      manji[x].push_back(manji[n][j]);
    }
    manji[x].push_back(n);
  }
  return;
}
void solve(int x) {
  for (int i = 0; i < graf[x].size(); i++) {
    int n = graf[x][i];
    solve(n);
    for (int j = 0; j < manji[n].size(); j++) {
      rj[x].push_back(make_pair(val[manji[n][j]], manji[n][j]));
    }
    rj[x].push_back(make_pair(val[n], n));
  }
  sort(rj[x].begin(), rj[x].end());
  if (M[x] == 0 and !rj[x].empty()) val[x] = val[rj[x][0].second] + 1;
  for (int i = 0; i < rj[x].size(); i++) {
    if (i + 1 == M[x]) val[x] = rj[x][i].first + 1;
    if (i + 1 > M[x]) val[rj[x][i].second] += 2;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> a >> b;
    if (a)
      graf[a].push_back(i);
    else
      root = i;
    M[i] = b;
  }
  dfs(root);
  bool da = 1;
  for (int i = 1; i <= N; i++) {
    if (manji[i].size() < M[i]) da = 0;
  }
  if (!da) {
    cout << "NO" << '\n';
    return 0;
  }
  solve(root);
  cout << "YES" << '\n';
  for (int i = 1; i <= N; i++) cout << val[i] << ' ';
}
