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;
void yes() {
  cout << "YES"
       << "\n";
}
void no() {
  cout << "NO"
       << "\n";
}
long long int ans[2005];
long long int vis[2005];
long long int vis1[2005];
vector<long long int> v[2005];
vector<long long int> v2[2005];
long long int tot[2005];
long long int tot1[2005];
long long int c[2005];
long long int cnt = 0;
long long int cntt = 0;
queue<long long int> st;
void dfs1(long long int x) {
  for (long long int i = 0; i < v2[x].size(); i++) {
    if (!vis1[v2[x][i]]) {
      vis1[v2[x][i]] = 1;
      dfs1(v2[x][i]);
    }
  }
  long long int anss = 0;
  for (long long int ii = 0; ii < v2[x].size(); ii++) {
    anss += tot1[v2[x][ii]];
  }
  tot[x] = anss;
  tot1[x] = anss + 1;
}
void dfs(long long int x) {
  for (long long int i = 0; i < v2[x].size(); i++) {
    if (!vis[v2[x][i]]) {
      vis[v2[x][i]] = 1;
      dfs(v2[x][i]);
    }
  }
  st.push(x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    long long int x, y;
    cin >> x >> y;
    v[i].push_back(x);
    v2[x].push_back(i);
    c[i] = y;
  }
  long long int ind = 0;
  for (long long int i = 0; i < n; i++) {
    if (v[i].size() != 0) {
      if (v[i][0] == 0) {
        ind = i;
        break;
      }
    }
  }
  vis1[ind] = 1;
  dfs1(ind);
  long long int flag = 0;
  for (long long int i = 1; i <= n; i++) {
    if (tot[i] < c[i]) {
      flag = 1;
      break;
    }
  }
  if (flag == 1) {
    no();
    exit(0);
  }
  vis[ind] = 1;
  dfs(ind);
  while (!st.empty()) {
    long long int x = st.front();
    st.pop();
    long long int in = c[x];
    for (long long int j = cnt; j >= in; j--) {
      ans[j + 1] = ans[j];
    }
    ans[in] = x;
    cnt++;
  }
  yes();
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 0; j < n + 1; j++) {
      if (ans[j] == i) {
        cout << j + 1 << " ";
        break;
      }
    }
  }
  cout << "\n";
  return 0;
}
