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 n;
vector<int> Graph[2111];
int c[2111];
int root;
vector<int> subtree[2111];
bool BAD = false;
void DFS(int ver, int dad) {
  int i, j;
  for (i = 0; i < Graph[ver].size(); i++) {
    if (Graph[ver][i] == dad) continue;
    DFS(Graph[ver][i], ver);
    if (BAD) return;
    for (j = 0; j < subtree[Graph[ver][i]].size(); j++) {
      subtree[ver].push_back(subtree[Graph[ver][i]][j]);
    }
  }
  if (c[ver] > subtree[ver].size()) {
    BAD = true;
    return;
  }
  subtree[ver].push_back(ver);
  int cur = subtree[ver].size() - 1;
  while (cur > c[ver]) {
    int swp = subtree[ver][cur];
    subtree[ver][cur] = subtree[ver][cur - 1];
    subtree[ver][cur - 1] = swp;
    cur--;
  }
}
int values[100111];
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    int p;
    scanf("%d %d", &p, &c[i]);
    if (p == 0) {
      root = i;
      continue;
    }
    Graph[p].push_back(i);
    Graph[i].push_back(p);
  }
  DFS(root, 0);
  if (BAD) {
    printf("NO\n");
    return 0;
  }
  for (i = 0; i < subtree[root].size(); i++) {
    values[subtree[root][i]] = i + 1;
  }
  printf("YES\n");
  for (i = 1; i <= n; i++) {
    printf("%d", values[i]);
    if (i != n) printf(" ");
  }
  printf("\n");
  return 0;
}
