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 _main();
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  _main();
}
const int inf = INT_MAX / 2;
const long long infl = 1LL << 60;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int N;
vector<int> E[2010];
int C[2010];
int ans[2010];
void dfs2(int cu, int lim) {
  if (lim <= ans[cu]) ans[cu]++;
  for (auto& to : E[cu]) dfs2(to, lim);
}
int dfs(int cu, int base = 0) {
  int cnt = 0;
  for (auto& to : E[cu]) cnt += dfs(to, base + cnt);
  if (cnt < C[cu]) {
    cout << "NO" << endl;
    exit(0);
  }
  for (auto& to : E[cu]) dfs2(to, base + C[cu] + 1);
  ans[cu] = base + C[cu] + 1;
  return cnt + 1;
}
void _main() {
  int root;
  cin >> N;
  for (int i = 0; i < N; i++) {
    int p;
    cin >> p >> C[i];
    p--;
    if (p < 0)
      root = i;
    else
      E[p].push_back(i);
  }
  dfs(root);
  printf("YES\n");
  for (int i = 0; i < N; i++) {
    if (i) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
}
