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 int MAXN = 2050;
int n;
vector<int> g[MAXN];
int root;
int c[MAXN];
int ans[MAXN];
int sz[MAXN];
int f[MAXN];
void update(int pos, int delta) {
  for (; pos < n; pos = (pos | (pos + 1))) {
    f[pos] += delta;
  }
}
long long sum(int pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    res += f[pos];
  }
  return res;
}
long long sum(int l, int r) { return sum(r) - sum(l - 1); }
int dfs(int v) {
  sz[v]++;
  for (int to : g[v]) {
    sz[v] += dfs(to);
  }
  return sz[v];
}
int getKth(int k) {
  int l = 1, r = n;
  while (l < r) {
    int mid = (l + r) / 2;
    if (sum(1, mid) < k) {
      l = mid + 1;
    } else {
      r = mid;
    }
  }
  return l;
}
void dfs2(int v) {
  if (c[v] > sz[v] - 1) {
    puts("NO");
    exit(0);
  }
  ans[v] = getKth(c[v] + 1);
  update(ans[v], -1);
  for (int to : g[v]) {
    dfs2(to);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int p;
    scanf("%d %d", &p, &c[i]);
    if (p == 0) {
      root = i;
    } else {
      p--;
      g[p].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    update(i, 1);
  }
  dfs(root);
  dfs2(root);
  puts("YES");
  for (int i = 0; i < n; i++) {
    printf("%d ", ans[i]);
  }
  cout << endl;
  return 0;
}
