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>
#pragma GCC optimize("-Ofast", "-funroll-all-loops")
using namespace std;
const long long N = 2e3 + 10;
long long n, res[N], p[N], c[N], rt, sz[N], vis[N];
long long head[N], nex[N], to[N], tot;
inline void add(long long a, long long b) {
  to[++tot] = b;
  nex[tot] = head[a];
  head[a] = tot;
}
void dfs_sz(long long x) {
  sz[x] = 1;
  for (long long i = head[x]; i; i = nex[i]) {
    dfs_sz(to[i]);
    sz[x] += sz[to[i]];
  }
}
void dfs(long long x) {
  long long cnt = 0;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) cnt++;
    if (cnt == c[x] + 1) {
      res[x] = i;
      break;
    }
  }
  vis[res[x]] = 1;
  for (long long i = head[x]; i; i = nex[i]) dfs(to[i]);
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    if (!p[i]) rt = i;
    if (p[i]) add(p[i], i);
  }
  dfs_sz(rt);
  for (long long i = 1; i <= n; i++)
    if (c[i] >= sz[i]) return puts("NO"), 0;
  dfs(rt);
  puts("YES");
  for (long long i = 1; i <= n; i++) cout << res[i] << ' ';
  puts("");
  return 0;
}
