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;
string vow = "aeiou";
int month[] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dxhorse[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dyhorse[] = {1, -1, 2, -2, 2, -2, 1, -1};
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
const long double pie = 3.1415926535897932384626;
const long long mod = 1e9 + 7;
const int N = 2e3 + 5;
vector<int> g[N];
int c[N];
int n, root;
void read() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p, cc;
    cin >> p >> cc;
    c[i] = cc;
    p--;
    if (p != -1) {
      g[p].push_back(i);
      g[i].push_back(p);
    } else
      root = i;
  }
}
int sub[N];
bool vis[N];
vector<int> order[N];
void dfs(int node) {
  vis[node] = 1;
  sub[node]++;
  for (auto i : g[node]) {
    if (vis[i]) continue;
    ;
    dfs(i);
    sub[node] += sub[i];
    for (auto ii : order[i]) order[node].push_back(ii);
  }
  if (c[node] > sub[node] - 1) {
    cout << "NO";
    cout << "\n";
    ;
    exit(0);
  }
  order[node].insert(order[node].begin() + c[node], node);
}
void solve(int test_case) {
  read();
  dfs(root);
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[order[root][i]] = i;
  cout << "YES";
  cout << "\n";
  ;
  for (auto i : ans) cout << i + 1 << " ";
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) solve(i);
  return 0;
}
