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 int MOD = 1000000007;
long long int modpow(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    res %= mod;
    a = a * a;
    a %= mod;
  }
  return res;
}
long long int p(long long int a, long long int b) {
  long long int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
void dfs(long long int node, long long int d, vector<vector<long long int>>& b,
         vector<long long int>& depth) {
  if (node == -1) return;
  depth[node] = d;
  for (long long int i : b[node]) dfs(i, d + 1, b, depth);
}
void testcases() {
  long long int n;
  cin >> n;
  vector<long long int> g(n);
  vector<vector<long long int>> b(n);
  vector<long long int> c(n);
  long long int root = -1;
  for (long long int i = (long long int)0; i < (long long int)n; ++i) {
    long long int x, y;
    cin >> x >> y;
    x--;
    g[i] = x;
    if (x != -1)
      b[x].push_back(i);
    else
      root = i;
    c[i] = y;
  }
  vector<long long int> depth(n);
  dfs(root, 0, b, depth);
  long long int flag = 1;
  set<tuple<long long int, long long int, long long int>> q;
  for (long long int i = (long long int)0; i < (long long int)n; ++i)
    q.insert({c[i], depth[i], i});
  vector<long long int> ans(n);
  long long int cnt = 1;
  while (!q.empty()) {
    tuple<long long int, long long int, long long int> i = *q.begin();
    q.erase(q.begin());
    if (get<0>(i) > 0) {
      flag = 0;
      break;
    }
    long long int cur = get<2>(i);
    ans[cur] = cnt++;
    cur = g[cur];
    while (cur != -1) {
      if (q.find({c[cur], depth[cur], cur}) != q.end()) {
        q.erase(q.find({c[cur], depth[cur], cur}));
        c[cur]--;
        q.insert({c[cur], depth[cur], cur});
      }
      cur = g[cur];
    }
  }
  if (!flag) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  for (long long int i : ans) cout << i << " ";
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  t = 1;
  while (t--) testcases();
}
