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 p[2005], c[2005];
int n, root = -1;
vector<int> v[2005];
int d[2005], vis[2005];
int ans[2005];
struct cmp {
  bool operator()(int x, int y) { return d[x] > d[y]; }
};
priority_queue<int, vector<int>, cmp> q;
void dfs(int x) {
  if (c[x] == 0) q.push(x);
  for (auto y : v[x]) {
    if (vis[y]) continue;
    vis[y] = 1;
    d[y] = d[x] + 1;
    dfs(y);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    if (!p[i]) {
      root = i;
      continue;
    }
    v[i].push_back(p[i]);
    v[p[i]].push_back(i);
  }
  vis[root] = 1;
  dfs(root);
  bool flag = true;
  int count = 1;
  while (!q.empty()) {
    int _top = q.top();
    q.pop();
    ans[_top] = count++;
    int head = p[_top];
    while (head) {
      c[head]--;
      if (c[head] == 0) q.push(head);
      head = p[head];
    }
  }
  for (int i = 1; i <= n; i++) {
    if (c[i] > 0) {
      flag = false;
      break;
    }
  }
  if (flag) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
