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;
vector<int> g[2005], ans;
int flag = 0;
int res[2005], vis[2005], v[2005], son[2005];
void find(int x) {
  vis[x] = 1;
  for (int i = 0; i < g[x].size(); i++) {
    int t = g[x][i];
    if (vis[t]) continue;
    find(t);
    son[x] += son[t] + 1;
  }
}
void dfs(int x) {
  if (flag) return;
  vis[x] = 1;
  if (son[x] < v[x]) {
    flag = 1;
    return;
  }
  res[x] = ans[v[x]];
  ans.erase(ans.begin() + v[x]);
  for (int i = 0; i < g[x].size(); i++) {
    int t = g[x][i];
    if (vis[t]) continue;
    dfs(t);
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int root;
  for (int i = 1; i <= n; i++) {
    ans.push_back(i);
    int x;
    cin >> x >> v[i];
    if (x == 0)
      root = i;
    else {
      g[x].push_back(i);
      g[i].push_back(x);
    }
  }
  find(root);
  memset(vis, 0, sizeof(vis));
  dfs(root);
  if (flag)
    cout << "NO" << '\n';
  else {
    cout << "YES" << '\n';
    for (int i = 1; i <= n; i++) {
      cout << res[i];
      if (i == n)
        cout << '\n';
      else
        cout << ' ';
    }
  }
  return 0;
}
