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;
const int MAXN = 2005;
int n;
int c[MAXN];
vector<int> graph[MAXN];
int answer[MAXN];
int treeSize[MAXN];
vector<int> dfsOrder;
void DFS(int x) {
  treeSize[x] = 1;
  dfsOrder.push_back(x);
  for (int y : graph[x]) {
    DFS(y);
    treeSize[x] += treeSize[y];
  }
}
bool active[MAXN];
int getKth(int k) {
  for (int i = 1; i <= n; i++) {
    if (active[i] == false) continue;
    if (k == 0) return i;
    k--;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int root;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int par;
    cin >> par >> c[i];
    if (par == 0)
      root = i;
    else
      graph[par].push_back(i);
  }
  DFS(root);
  for (int i = 1; i <= n; i++) active[i] = true;
  for (int x = 1; x <= n; x++) {
    if (treeSize[x] - 1 < c[x]) {
      cout << "NO" << '\n';
      return 0;
    }
  }
  for (int x : dfsOrder) {
    answer[x] = getKth(c[x]);
    active[answer[x]] = false;
  }
  cout << "YES" << '\n';
  for (int x = 1; x <= n; x++) {
    cout << answer[x] << " ";
  }
  cout << '\n';
}
