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 to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) res += ", ";
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  dbg_out(T...);
}
pair<long long, long long> dir[8] = {{-1, 0}, {1, 0},   {0, -1}, {0, 1},
                                     {1, 1},  {-1, -1}, {-1, 1}, {1, -1}};
const long long inf = (1LL << 60), mod = 1e9 + 7, nn = 2003;
long long n, c[nn], a[nn], sz[nn];
vector<long long> g[nn];
vector<long long> order[nn];
void add_child_to_parent(vector<long long> &child, vector<long long> &par) {
  for (long long i = 0; i < child.size(); i++) {
    par.push_back(child[i]);
  }
}
void dfs(long long v, long long par) {
  sz[v] = 1;
  for (long long child : g[v]) {
    if (child != par) {
      dfs(child, v);
      add_child_to_parent(order[child], order[v]);
      sz[v] += sz[child];
    }
  }
  if (c[v] >= sz[v]) {
    cout << "NO";
    exit(0);
  }
  order[v].insert(order[v].begin() + c[v], v);
}
void solve() {
  cin >> n;
  long long p, root;
  for (long long i = 1; i <= n; ++i) {
    cin >> p >> c[i];
    if (p) {
      g[p].push_back(i);
      g[i].push_back(p);
    } else
      root = i;
  }
  dfs(root, 0);
  for (long long i = 0; i < order[root].size(); ++i) {
    a[order[root][i]] = (i + 1);
  }
  cout << "YES\n";
  for (long long i = 1; i <= n; i++) cout << a[i] << ' ';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  for (long long tc = 1; tc <= T; tc++) {
    solve();
  }
  return 0;
}
