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 = 2e3 + 10;
const int mod = 1e9 + 7;
int c[maxn], vis[maxn];
vector<int> E[maxn];
int sz[maxn], ans[maxn];
int f, n;
int fk(int x) {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    cnt++;
    if (cnt == x) {
      vis[i] = 1;
      return i;
    }
  }
  return 0;
}
void dfs(int u) {
  sz[u] = 1;
  int x = fk(c[u] + 1);
  if (!x) f = 1;
  ans[u] = x;
  for (auto &v : E[u]) {
    dfs(v);
    sz[u] += sz[v];
  }
  if (sz[u] - 1 < c[u]) f = 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int rt = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x >> c[i];
    if (!x)
      rt = i;
    else
      E[x].push_back(i);
  }
  dfs(rt);
  if (f)
    cout << "NO" << '\n';
  else {
    cout << "YES" << '\n';
    for (int i = 1; i <= n; i++) {
      if (i != 1) cout << " ";
      cout << ans[i];
    }
  }
}
