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 N = 3e5 + 10;
const int M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
int n, k;
vector<int> G[N];
int a[N], c[N], sz[N], ans[N];
void dfs1(int u) {
  sz[u] = 1;
  for (auto &v : G[u]) {
    dfs1(v);
    sz[u] += sz[v];
  }
}
void add(int x, int v) {
  for (; x <= n; x += (x & (-x))) c[x] += v;
}
int ask(int x) {
  int res = 0;
  for (; x > 0; x -= (x & (-x))) res += c[x];
  return res;
}
int k_th(int k) {
  int l = 1, r = n;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (ask(mid) >= k)
      r = mid;
    else
      l = mid + 1;
  }
  add(l, -1);
  return l;
}
void dfs2(int u) {
  if (a[u] >= sz[u]) {
    puts("NO");
    exit(0);
  }
  ans[u] = k_th(a[u] + 1);
  for (auto &v : G[u]) dfs2(v);
}
int main() {
  scanf("%d", &n);
  int root, p;
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &p, &a[i]);
    if (!p)
      root = i;
    else
      G[p].push_back(i);
    add(i, 1);
  }
  dfs1(root);
  dfs2(root);
  puts("YES");
  for (int i = 1; i <= n; ++i) {
    printf("%d ", ans[i]);
  }
  return 0;
}
