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;
vector<int> g[2005], vals[2005];
int p[2005], c[2005], sons[2005], res[2005];
bool ok;
void dfs(int u) {
  sons[u] = 1;
  for (auto v : g[u]) {
    dfs(v);
    sons[u] += sons[v];
  }
}
void dfs1(int u) {
  if (c[u] >= sons[u]) {
    ok = 0;
    return;
  }
  res[u] = vals[u][c[u]];
  queue<int> ss;
  for (int i = 0; i < sons[u]; i++)
    if (i != c[u]) ss.push(vals[u][i]);
  for (auto v : g[u]) {
    for (int i = 0; i < sons[v]; i++) {
      vals[v].push_back(ss.front());
      ss.pop();
    }
    dfs1(v);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int root = -1;
  for (int i = 0; i < n; i++) {
    cin >> p[i] >> c[i];
    if (!p[i])
      root = i;
    else {
      p[i]--;
      g[p[i]].push_back(i);
    }
  }
  dfs(root);
  for (int i = 1; i <= n; i++) vals[root].push_back(i);
  ok = 1;
  dfs1(root);
  if (ok) {
    cout << ("YES") << '\n';
    for (int i = 0; i < n; i++) cout << res[i] << ' ';
    cout << '\n';
  } else
    cout << ("NO") << '\n';
  return 0;
}
