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(3)
using namespace std;
const int Mod = 1e9 + 7;
const int N = 2e3 + 100;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
inline long long dpow(long long a, long long b) {
  long long r = 1, t = a;
  while (b) {
    if (b & 1) r = (r * t) % Mod;
    b >>= 1;
    t = (t * t) % Mod;
  }
  return r;
}
inline long long fpow(long long a, long long b) {
  long long r = 1, t = a;
  while (b) {
    if (b & 1) r = (r * t);
    b >>= 1;
    t = (t * t);
  }
  return r;
}
vector<int> G[N], vec;
int a[N], sz[N], n;
int ans[N];
void DFS(int x) {
  sz[x] = 1;
  for (auto v : G[x]) {
    DFS(v);
    sz[x] += sz[v];
  }
  if (sz[x] - 1 < a[x]) puts("NO"), exit(0);
}
void DFS1(int x) {
  ans[x] = vec[a[x] + 1];
  vec.erase(vec.begin() + a[x] + 1);
  for (auto v : G[x]) DFS1(v);
}
int main() {
  cin >> n;
  vec.push_back(0);
  int rt = 1;
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d %d", &u, &a[i]);
    if (u)
      G[u].push_back(i);
    else
      rt = i;
    vec.push_back(i);
  }
  DFS(rt);
  DFS1(rt);
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
