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;
vector<int> g[2010];
int c[2010], lzy[2010], son[2010] = {0}, ans[2010], mxn[2010], flag = 1,
                        vis[2010];
void dfs2(int x) {
  if (!flag) return;
  int cnt = 0;
  for (int j = 1; j <= n; j++) {
    if (!vis[j]) cnt++;
    if (cnt == c[x] + 1) {
      ans[x] = j;
      vis[j] = 1;
      break;
    }
  }
  for (int i = 0; i < g[x].size(); i++) {
    int v = g[x][i];
    dfs2(v);
    if (!flag) return;
    son[x] += son[v] + 1;
  }
  if (c[x] > son[x]) flag = 0;
}
int main() {
  cin >> n;
  int r;
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d%d", &p, &c[i]);
    if (p)
      g[p].push_back(i);
    else
      r = i;
  }
  dfs2(r);
  if (!flag) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
}
