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;
int c[2005];
int a[2005];
int parent[2005];
int indegree[2005];
int cnt[2005];
queue<int> q;
vector<vector<int>> children(2005);
vector<int> v;
bool pos;
void solve2(int x, vector<int> &v) {
  if ((int)v.size() <= c[x]) {
    pos = false;
    return;
  }
  a[x] = v[c[x]];
  if ((int)v.size() == 1) return;
  v.erase(v.begin() + c[x]);
  vector<int> tmp;
  int num = 0;
  for (auto child : children[x]) {
    tmp.clear();
    for (int i = 0; i <= cnt[child]; i++) {
      tmp.push_back(v[num + i]);
    }
    num += cnt[child] + 1;
    solve2(child, tmp);
  }
}
void solve() {
  int n;
  int root;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &parent[i], &c[i]);
    if (parent[i] != 0) {
      children[parent[i]].push_back(i);
      indegree[parent[i]]++;
    } else
      root = i;
  }
  for (int i = 1; i <= n; i++)
    if (indegree[i] == 0) q.push(i);
  while (!q.empty()) {
    int here = q.front();
    q.pop();
    cnt[parent[here]] += cnt[here] + 1;
    indegree[parent[here]]--;
    if (indegree[parent[here]] == 0) q.push(parent[here]);
  }
  for (int i = 1; i <= n; i++) v.push_back(i);
  solve2(root, v);
  if (pos) {
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  } else
    printf("NO\n");
}
int main(void) {
  pos = true;
  solve();
  return 0;
}
