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>
#pragma GCC optimize("O2")
using namespace std;
const long long int mod = 1000000007;
const long long int INF64 = mod * mod + 1;
class segTree {
  long long int sz;
  vector<long long int> val;
  vector<long long int> opt;
  vector<long long int> len;
  long long int NOP = 0;
  long long int NEU = 0;

 public:
  segTree(long long int n) {
    sz = 1;
    while (sz < n) sz *= 2;
    val.assign(2 * sz, NEU);
    opt.assign(2 * sz, 0LL);
    len.assign(2 * sz, 0LL);
  }
  void updateIndex(long long int i, long long int v, long long int x,
                   long long int lx, long long int rx) {
    if (rx == lx) {
      val[x] += v;
      return;
    }
    long long int m = (lx + rx) / 2;
    if (i <= m)
      updateIndex(i, v, 2 * x + 1, lx, m);
    else
      updateIndex(i, v, 2 * x + 2, m + 1, rx);
    val[x] = val[2 * x + 1] + val[2 * x + 2];
  }
  void updateIndex(long long int i, long long int v) {
    updateIndex(i, v, 0, 0, sz - 1);
  }
  long long int query(long long int k, long long int x, long long int lx,
                      long long int rx) {
    if (rx == lx) return lx;
    long long int m = (lx + rx) / 2;
    if (val[2 * x + 1] >= k)
      return query(k, 2 * x + 1, lx, m);
    else
      return query(k - val[2 * x + 1], 2 * x + 2, m + 1, rx);
  }
  long long int query(long long int k) { return query(k, 0, 0, sz - 1); }
};
void solve() {
  long long int n;
  cin >> n;
  long long int root;
  vector<long long int> c(n + 1);
  vector<long long int> adj[n + 1];
  for (long long int i = 1; i <= n; i++) {
    long long int p;
    cin >> p >> c[i];
    if (!p) root = i;
    adj[p].push_back(i);
  }
  bool invalid = false;
  function<long long int(long long int)> dfs1 = [&](long long int i) {
    long long int subtree = 0;
    for (auto it : adj[i]) subtree += dfs1(it);
    if (c[i] > subtree) invalid = true;
    return subtree + 1;
  };
  dfs1(root);
  if (invalid) {
    cout << "NO" << '\n';
    return;
  }
  segTree st(n);
  for (long long int i = 0; i < n; i++) st.updateIndex(i, 1);
  vector<long long int> ans(n + 1, 0);
  function<void(long long int)> dfs2 = [&](long long int i) {
    ans[i] = st.query(c[i] + 1) + 1;
    st.updateIndex(ans[i] - 1, -1);
    for (auto it : adj[i]) dfs2(it);
  };
  dfs2(root);
  cout << "YES" << '\n';
  for (long long int i = 1; i <= n; i++) cout << ans[i] << " ";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  for (long long int i = 1; i <= T; i++) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
