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 kMaxN = 2007;
int n, p[kMaxN], c[kMaxN], sizes[kMaxN], a[kMaxN], root;
vector<int> graph[kMaxN];
struct Node {
  int value;
  Node *left;
  Node *right;
  Node(int v = 0, Node *l = nullptr, Node *r = nullptr) {
    value = v;
    left = l;
    right = r;
  }
};
typedef Node *Tree;
Tree the_tree;
Tree build_tree(int low = 0, int high = n) {
  if (low + 1 == high) {
    return new Node(1);
  }
  int mid = (low + high) >> 1;
  Tree l = build_tree(low, mid);
  Tree r = build_tree(mid, high);
  return new Node(l->value + r->value, l, r);
}
int get(int k, Tree t = the_tree, int low = 0, int high = n) {
  t->value--;
  if (low + 1 == high) {
    return low;
  }
  int mid = (low + high) >> 1;
  if (k <= t->left->value) {
    return get(k, t->left, low, mid);
  }
  return get(k - t->left->value, t->right, mid, high);
}
void dfs_2(int u = root) {
  ;
  a[u] = get(c[u] + 1);
  for (int v : graph[u]) dfs_2(v);
}
bool dfs_1(int u = root) {
  sizes[u] = 1;
  for (int v : graph[u]) {
    if (!dfs_1(v)) return false;
    sizes[u] += sizes[v];
  }
  if (sizes[u] - 1 < c[u]) return false;
  return true;
}
void solve() {
  cin >> n;
  for (int u = 0; u < n; u++) {
    cin >> p[u] >> c[u];
    p[u]--;
    if (p[u] != -1)
      graph[p[u]].emplace_back(u);
    else
      root = u;
  }
  if (!dfs_1()) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  the_tree = build_tree();
  dfs_2();
  for (int u = 0; u < n; u++) {
    cout << a[u] + 1 << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int test_cases_num = 1;
  for (int i = 1; i <= test_cases_num; i++) {
    if (false)
      if (test_cases_num > 1)
        if (false) cerr << "Test #" << i << ":\n";
    solve();
    if (false)
      if (test_cases_num > 1)
        if (false) cerr << "End of test #" << i << ".\n";
  }
  return 0;
}
