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 = 2005;
int n;
int c[maxn];
int root = 1;
vector<int> g[maxn];
int ans[maxn];
void ReadInput() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p >> c[i];
    if (p == 0)
      root = i;
    else {
      g[p].push_back(i);
      g[i].push_back(p);
    }
  }
}
int sz[maxn];
void dfs1(int u, int par) {
  sz[u] = 1;
  for (int v : g[u])
    if (v != par) {
      dfs1(v, u);
      sz[u] += sz[v];
    }
}
vector<int> order[maxn];
void dfs2(int u, int par) {
  for (int v : g[u])
    if (v != par) {
      dfs2(v, u);
      for (int it : order[v]) order[u].push_back(it);
    }
  order[u].insert(order[u].begin() + c[u], u);
}
void Solve() {
  dfs1(root, 0);
  for (int i = 1; i <= n; ++i)
    if (sz[i] - 1 < c[i]) {
      cout << "NO\n";
      return;
    }
  dfs2(root, 0);
  cout << "YES\n";
  for (int i = 0; i < n; ++i) ans[order[root][i]] = i;
  for (int i = 1; i <= n; ++i) cout << ans[i] + 1 << ' ';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ReadInput();
  Solve();
}
