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 = 2020;
int n;
int cnt;
int child[MAXN];
int L[MAXN], R[MAXN];
int inv[MAXN];
int ans[MAXN];
int order[MAXN];
vector<int> g[MAXN];
void upd(int x, int delta, int bound) {
  for (int id = L[x]; id <= R[x]; ++id) {
    int v = inv[id];
    if (ans[v] >= bound) ans[v] += delta;
  }
}
int sz;
int a[MAXN];
int find(int x, int kth) {
  sz = 0;
  for (int id = L[x]; id <= R[x]; ++id) {
    int v = inv[id];
    a[++sz] = ans[v];
  }
  sort(a + 1, a + sz + 1);
  return a[kth];
}
void write(int x) {
  for (int id = L[x]; id <= R[x]; ++id) {
    int v = inv[id];
    cout << ans[v] << " ";
  }
  cout << endl;
}
void solve(int x) {
  L[x] = ++cnt;
  inv[cnt] = x;
  int temp = order[x];
  for (int v : g[x]) {
    solve(v);
    upd(v, child[x], 0);
    child[x] += child[v];
    if (child[v] >= temp) {
      int val = find(v, temp) + 1;
      upd(v, 1, val);
      ans[x] = val;
      child[x]++;
      temp = 1e9;
    } else {
      temp -= child[v];
    }
  }
  if (temp <= n) {
    if (temp != 0) {
      cout << "NO";
      exit(0);
    }
    child[x]++;
    ans[x] = child[x];
  }
  R[x] = cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p >> order[i];
    g[p].push_back(i);
  }
  solve(0);
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " ";
  }
  return 0;
}
