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 MN = 2e3 + 2, MOD = 1e9 + 7, REEE = 100;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
vector<int> adj[MN];
int c[MN];
int a[MN];
int n;
int p[MN];
int order[MN], st[MN], ed[MN], ree = 0;
int o[MN];
int hmm = 0;
bool fail = 0;
void dfs(int cur) {
  int lb = 1, ub = 1e9;
  int temp = cur;
  while (p[temp]) {
    temp = p[temp];
    if (c[temp] < c[cur] + 1)
      lb = max(lb, a[temp]);
    else
      ub = min(ub, a[temp]);
  }
  a[cur] = lb + ub >> 1;
  temp = cur;
  while (p[temp]) {
    temp = p[temp];
    if (a[cur] < a[temp]) --c[temp];
  }
  for (int i : adj[cur]) dfs(i);
}
void getorder(int cur) {
  st[cur] = ++ree;
  order[ree] = cur;
  for (int i : adj[cur]) getorder(i);
  ed[cur] = ree;
}
int main() {
  int root = -1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &p[i], &o[i]);
    c[i] = o[i];
    if (p[i])
      adj[p[i]].push_back(i);
    else
      root = i;
  }
  getorder(root);
  dfs(root);
  for (int cur = 1; cur <= n; cur++) {
    int cnt = 0;
    for (int i = st[cur] + 1; i <= ed[cur]; i++) {
      if (a[order[i]] < a[cur]) ++cnt;
    }
    if (cnt != o[cur]) {
      return 0 * printf("NO\n");
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  printf("\n");
  return 0;
}
