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 long long inf = 1e9 + 5;
const long long INF = 1e18 + 5;
const long long maxn = 1e6 + 5;
long long n, a[maxn], b[maxn];
long long w[maxn], dp[maxn];
vector<int> g[maxn];
long long ans = 0;
vector<int> vv[maxn];
void dfs(int v) {
  if (b[v] == 0) vv[v].push_back(v);
  for (auto to : g[v]) {
    dfs(to);
    for (auto i : vv[to]) {
      vv[v].push_back(i);
      if (vv[v].size() == b[v]) vv[v].push_back(v);
    }
  }
  if (vv[v].size() < b[v]) {
    ans = 1;
    return;
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(NULL);
  long long n;
  cin >> n;
  int x = 0;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> b[i];
    if (p == 0)
      x = i;
    else
      g[p].push_back(i);
  }
  dfs(x);
  if (ans == 1) return cout << "NO", 0;
  for (int i = 0; i < vv[x].size(); i++) dp[vv[x][i]] = i + 1;
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) cout << dp[i] << ' ';
  return 0;
}
