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 = 2010;
const int inf = 1e9 + 1e8;
int N, par[MAXN], c[MAXN];
vector<int> graph[MAXN];
bool works = true;
int ans[MAXN], sz[MAXN], root, solved;
int get_min(int cur) {
  int min_val = ans[cur];
  for (int node : graph[cur]) {
    min_val = min(min_val, get_min(node));
  }
  return min_val;
}
void dfs(int cur, int rev) {
  if (ans[cur] >= rev) {
    ans[cur]++;
  }
  for (int node : graph[cur]) {
    dfs(node, rev);
  }
  return;
}
void solve(int cur) {
  sz[cur] = 1;
  if (graph[cur].size() == 0) {
    if (c[cur] >= sz[cur]) {
      works = false;
      return;
    }
    ans[cur] = c[cur] + solved + 1;
    solved++;
    return;
  }
  for (int node : graph[cur]) {
    solve(node);
    sz[cur] += sz[node];
  }
  if (c[cur] >= sz[cur]) {
    works = false;
    return;
  }
  ans[cur] = inf;
  int minn = get_min(cur);
  dfs(cur, minn + c[cur]);
  ans[cur] = minn + c[cur];
  solved++;
  return;
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> par[i] >> c[i];
    if (par[i] == 0) {
      root = i;
      continue;
    }
    graph[par[i]].push_back(i);
  }
  solve(root);
  if (!works) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= N; i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
  return 0;
}
