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<vector<int> > children;
vector<int> c, a;
bool possible = true;
int nextVal = 1;
void changeVals(int node, int barrier) {
  if (a[node] >= barrier) ++a[node];
  for (int child : children[node]) changeVals(child, barrier);
}
void findVals(int node, vector<int> &x) {
  x.push_back(a[node]);
  for (int child : children[node]) findVals(child, x);
}
void dfs(int node) {
  vector<int> subtree;
  for (int child : children[node]) {
    dfs(child);
    if (!possible) return;
    findVals(child, subtree);
  }
  sort(subtree.begin(), subtree.end());
  if (c[node] > subtree.size()) {
    possible = false;
    return;
  }
  if (c[node] == subtree.size())
    a[node] = nextVal;
  else {
    a[node] = subtree[c[node]];
    for (int child : children[node]) changeVals(child, a[node]);
  }
  ++nextVal;
}
int main() {
  int n;
  cin >> n;
  children.resize(1 + n), c.resize(1 + n), a.resize(1 + n);
  for (int i = 1; i <= n; ++i) {
    int pi;
    cin >> pi >> c[i];
    children[pi].push_back(i);
  }
  dfs(children[0][0]);
  if (possible) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
    cout << endl;
  } else
    cout << "NO" << endl;
  return 0;
}
