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;
template <class X, class Y>
bool minimize(X &x, const Y &y) {
  X eps = 1e-9;
  if (x > y + eps) {
    x = y;
    return true;
  } else
    return false;
}
template <class X, class Y>
bool maximize(X &x, const Y &y) {
  X eps = 1e-9;
  if (x + eps < y) {
    x = y;
    return true;
  } else
    return false;
}
template <class T>
T Abs(const T &x) {
  return (x < 0 ? -x : x);
}
const string YES = "YES";
const string NO = "NO";
void noAnswer(void) {
  cout << NO << endl;
  exit(EXIT_SUCCESS);
}
int numNode, root;
int par[3030], numLess[3030], res[3030];
vector<int> children[3030];
void loadTree(void) {
  scanf("%d", &numNode);
  for (int i = (1), _b = (numNode); i <= _b; i++) {
    scanf("%d%d", &par[i], &numLess[i]);
    if (par[i] == 0)
      root = i;
    else
      children[par[i]].push_back(i);
  }
}
vector<int> getNode(int u) {
  vector<int> nodes;
  for (__typeof((children[u]).begin()) it = (children[u]).begin();
       it != (children[u]).end(); it++) {
    vector<int> tmp = getNode(*it);
    for (__typeof((tmp).begin()) jt = (tmp).begin(); jt != (tmp).end(); jt++)
      nodes.push_back(*jt);
  }
  if (numLess[u] < 0 || numLess[u] > nodes.size()) noAnswer();
  nodes.insert(nodes.begin() + numLess[u], u);
  return nodes;
}
void process(void) {
  vector<int> nodes = getNode(root);
  for (int i = 0, _n = (nodes.size()); i < _n; i++) res[nodes[i]] = i + 1;
  cout << YES << endl;
  for (int i = (1), _b = (numNode); i <= _b; i++) printf("%d ", res[i]);
  printf("\n");
}
int main(void) {
  loadTree();
  process();
  return 0;
}
