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 N = 2005;
int n, a[N], par[N], root, ans[N], c;
vector<vector<int> > g;
void dfs2(int u, int val, bool b = 0) {
  if (!b) {
    if (ans[u] >= val) {
      ans[u]++;
    }
  }
  for (auto v : g[u]) {
    dfs2(v, val);
  }
}
set<int> dfs(int u) {
  set<int> st;
  for (auto v : g[u]) {
    set<int> t = dfs(v);
    for (auto x : t) {
      st.insert(x);
    }
  }
  if (st.empty()) {
    if (a[u] != 0) {
      cout << "NO\n";
      exit(0);
    }
    ans[u] = 10000 * (c++);
  } else {
    if (a[u] > (int)st.size()) {
      cout << "NO\n";
      exit(0);
    }
    int cnt = 0, val = -1;
    for (auto x : st) {
      if (cnt == a[u]) {
        val = x;
        break;
      }
      cnt++;
    }
    if (val == -1) {
      val = *st.rbegin();
      val++;
    }
    dfs2(u, val, 1);
    set<int> t;
    for (auto x : st) {
      if (x >= val) {
        t.insert(x + 1);
      } else {
        t.insert(x);
      }
    }
    st = t;
    ans[u] = val;
  }
  st.insert(ans[u]);
  return st;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  g.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> par[i] >> a[i];
    if (par[i]) {
      g[par[i] - 1].push_back(i);
    } else {
      root = i;
    }
  }
  dfs(root);
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << ans[i] + 1 << " ";
  }
}
