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;
int n;
int par[2000];
int c[2000];
vector<int> g[2000];
bool error = false;
int sz[2000];
int dfs(int root) {
  int ret = 0;
  for (int i = 0; i < g[root].size(); i++) {
    int x = dfs(g[root][i]);
    ret += x + 1;
  }
  sz[root] = ret;
  return ret;
}
int old = 0;
vector<pair<int, int> > solve(int root) {
  if (c[root] > sz[root]) {
    error = true;
  }
  if (g[root].size() == 0) {
    vector<pair<int, int> > ret;
    ret.push_back(make_pair(++old, root));
    return ret;
  }
  vector<pair<int, int> > ret;
  for (int i = 0; i < g[root].size(); i++) {
    vector<pair<int, int> > a = solve(g[root][i]);
    ret.insert(ret.end(), a.begin(), a.end());
  }
  int cnt = c[root];
  vector<pair<int, int> > final;
  for (int i = 0; i < cnt; i++) {
    final.push_back(ret[i]);
  }
  if (cnt == 0) {
    final.push_back(make_pair(1, root));
  } else
    final.push_back(make_pair(ret[cnt - 1].first + 1, root));
  int k = 1;
  for (int i = cnt; i < ret.size(); i++) {
    pair<int, int> p = ret[i];
    if (true)
      final.push_back(make_pair(p.first + 1, p.second));
    else
      final.push_back(make_pair(p.first, p.second));
    k++;
  }
  return final;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  int root = -1;
  for (int i = 0; i < n; i++) {
    cin >> par[i] >> c[i];
    if (par[i] != 0) g[par[i] - 1].push_back(i);
    if (par[i] == 0) root = i;
    par[i]--;
  }
  dfs(root);
  vector<pair<int, int> > ans = solve(root);
  if (!error) {
    int out[n];
    for (int i = 0; i < ans.size(); i++) {
      out[ans[i].second] = i + 1;
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      cout << out[i] << " ";
    }
  } else
    cout << "NO\n";
  return 0;
}
