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 = 2e3 + 10;
const long long INF = 1e9 + 10;
const int MOD = 1e18 + 7;
vector<int> tree[N];
int ans[N], cost[N], sz[N];
bool f = 1;
void dfs(int go, int cur, int n) {
  sz[go] = 1;
  for (auto u : tree[go]) {
    if (u != cur) {
      dfs(u, go, n);
      sz[go] += sz[u];
    }
  }
  if (cost[go] > sz[go] - 1) f = 0;
  cost[go]++;
  for (int i = 1; i <= n; i++) {
    if (ans[i] >= cost[go]) ans[i]++;
  }
  ans[go] = cost[go];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, root = -1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x, c;
    cin >> x >> c;
    if (x == 0)
      root = i;
    else
      tree[x].push_back(i);
    cost[i] = c;
  }
  dfs(root, 0, n);
  if (!f)
    cout << "NO" << '\n';
  else {
    cout << "YES" << '\n';
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  }
}
