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 = 2000 + 100;
int n;
vector<int> adj[maxn];
int c[maxn];
int a[maxn];
int idx[maxn];
int subsz[maxn];
int cur = maxn - 50;
bool func(int u) {
  bool ret = true;
  for (auto v : adj[u]) {
    ret = ret & func(v);
  }
  subsz[u]--;
  if (subsz[u] < c[u]) ret = false;
  if (ret) {
    int st = cur + c[u];
    for (int i = cur + 1; i <= st; i++) {
      a[idx[i]] = i - 1;
      idx[i - 1] = idx[i];
    }
    idx[st] = u;
    a[u] = st;
    cur--;
  }
  return ret;
}
void calculatesubsize(int u) {
  subsz[u] = 1;
  for (auto v : adj[u]) {
    calculatesubsize(v);
    subsz[u] += subsz[v];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d %d", &p, &c[i]);
    adj[p].push_back(i);
  }
  calculatesubsize(adj[0][0]);
  bool ans = func(adj[0][0]);
  if (ans) {
    printf("YES\n");
    for (int i = 1; i <= n; i++) {
      printf("%d ", a[i]);
    }
    printf("\n");
  } else {
    printf("NO\n");
  }
  return 0;
}
