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;
double pi = acos(-1);
long long power(long long a, long long b) {
  long long p_res = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      p_res *= a;
      b--;
    }
    a *= a;
    b /= 2;
  }
  return p_res;
}
long long gcd(long long a, long long b) {
  long long tmp;
  while (b != 0) {
    tmp = a;
    a = b;
    b = tmp % b;
  }
  return a;
}
const int MAXN = 2001;
int n;
int T[4 * MAXN];
int c[MAXN];
int sz[MAXN];
int ans[MAXN];
int rt;
int h = 1;
vector<int> adj[MAXN];
void update(int i, int v) {
  while (i >= 1) {
    T[i] += v;
    i >>= 1;
  }
}
int query(int i, int k) {
  if (i > h) return i - h;
  int res;
  int l = T[i * 2];
  if (k > l)
    res = query(i * 2 + 1, k - l);
  else
    res = query(i * 2, k);
  return res;
}
void dfs(int u) {
  ans[u] = query(1, c[u] + 1);
  update(h + ans[u], -1);
  for (auto v : adj[u]) {
    dfs(v);
  }
}
int cal_sub(int u) {
  for (auto v : adj[u]) {
    sz[u] += 1 + cal_sub(v);
  }
  return sz[u];
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> c[i];
    if (p == 0)
      rt = i;
    else
      adj[p].push_back(i);
  }
  cal_sub(rt);
  for (int i = 1; i <= n; i++) {
    if (sz[i] < c[i]) {
      cout << "NO";
      return 0;
    }
  }
  while (h < MAXN) h <<= 1;
  for (int i = 1; i <= n; i++) update(h + i, 1);
  dfs(rt);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
