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 N = 2e3 + 11;
vector<int> tree[N];
int c[N];
int dp[N];
int ans[N];
set<int> Set;
bool vis[N];
int n;
int dfs(int u, int par, int low) {
  int sum = 0;
  int cnt = 0;
  int idx = -1;
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    if (cnt == c[u]) {
      idx = i;
      break;
    }
    cnt++;
  }
  if (idx == -1) {
    cout << "NO";
    exit(0);
  }
  ans[u] = idx;
  vis[idx] = true;
  for (auto v : tree[u]) {
    if (v == par) continue;
    int val = dfs(v, u, sum);
    sum = sum + val;
  }
  if (sum < c[u]) {
    cout << "NO";
    exit(0);
  }
  return (sum + 1);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin >> n;
  int root;
  for (int i = 1; i <= n; ++i) {
    int par;
    cin >> par >> c[i];
    if (par == 0)
      root = i;
    else {
      tree[i].push_back(par);
      tree[par].push_back(i);
    }
  }
  for (int i = 1; i <= n; ++i) Set.insert(i);
  dfs(root, -1, 0);
  cout << "YES" << '\n';
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}
