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;
long long INF = 1e18;
vector<long long> g[2020];
vector<long long> ga[2020];
vector<long long> gb[2020];
long long c[2020];
long long d[2020];
long long vis[2020];
vector<pair<long long, long long>> res;
vector<pair<long long, long long>> inf;
void dfs(long long o, long long v) {
  if (vis[v]) return;
  vis[v] = 1;
  if (v != o) {
    ga[o].push_back(v);
    gb[v].push_back(o);
  }
  for (auto i : g[v]) dfs(o, i);
}
void dfs2(long long p, long long v) {
  if (vis[v]) return;
  vis[v] = 1;
  if (p != -1) d[v] = d[p] + 1;
  for (auto i : g[v]) dfs2(v, i);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long root = -1;
  for (int i = 0, ggdem = n; i < ggdem; ++i) {
    long long p;
    cin >> p;
    p--;
    if (p == -1)
      root = i;
    else
      g[p].push_back(i);
    cin >> c[i];
  }
  for (int i = 0, ggdem = n; i < ggdem; ++i) {
    memset((vis), (0), sizeof(vis));
    dfs(i, i);
  }
  memset((vis), (0), sizeof(vis));
  dfs2(-1, root);
  for (int i = 0, ggdem = n; i < ggdem; ++i) {
    inf.push_back({((int)ga[i].size()) - c[i], d[i]});
  }
  while (((int)res.size()) < n) {
    pair<long long, long long> va = {INF, INF};
    long long vai = -1;
    for (int i = 0, ggdem = n; i < ggdem; ++i) {
      if (va > inf[i]) {
        va = inf[i];
        vai = i;
      }
    }
    if (va.first != 0) {
      cout << "NO";
      return 0;
    }
    res.push_back({vai, n - ((int)res.size())});
    for (auto i : gb[vai]) {
      if (inf[i].first != INF) {
        inf[i].first--;
      }
    }
    inf[vai] = {INF, INF};
  }
  sort(res.begin(), res.end());
  cout << "YES\n";
  for (auto i : res) cout << i.second << " ";
  return 0;
}
