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;
using ll = long long;
using ld = long double;
using ii = pair<ll, ll>;
using vi = vector<ll>;
const ll N = 2e3 + 5;
ll n;
vector<vi> v(N);
ll cnt[N];
ll ans[N];
ll tree[N];
ll sz[N];
void upd(ll a, ll b) {
  while (a <= n) {
    tree[a] += b;
    a += a & -a;
  }
}
ll sum(ll a) {
  ll s = 0;
  while (a > 0) {
    s += tree[a];
    a -= a & -a;
  }
  return s;
}
ll src(ll a) {
  ll l = 1, r = n, ans = -1;
  while (l <= r) {
    ll mid = (l + r) / 2;
    if (sum(mid) >= a) {
      r = mid - 1;
      ans = mid;
    } else {
      l = mid + 1;
    }
  }
  return ans;
}
void dfs(ll a, ll b) {
  ans[a] = src(cnt[a] + 1);
  upd(ans[a], -1);
  for (auto u : v[a]) {
    if (u == b) continue;
    dfs(u, a);
  }
}
void findSz(ll a, ll b) {
  sz[a] = 1;
  for (auto u : v[a]) {
    if (u == b) continue;
    findSz(u, a);
    sz[a] += sz[u];
  }
  if (sz[a] <= cnt[a]) sz[a] = (ll)1e9;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  ll root;
  for (int i = 1; i <= n; i++) {
    ll d;
    cin >> d >> cnt[i];
    if (d == 0) {
      root = i;
      continue;
    }
    v[i].push_back(d);
    v[d].push_back(i);
  }
  for (int i = 1; i <= n; i++) upd(i, 1);
  findSz(root, 0);
  if (sz[root] >= (ll)1e9) {
    cout << "NO\n";
    return 0;
  }
  dfs(root, 0);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
