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 comment(linker, "/STACK:1000000000")
using namespace std;
const int maxn = 2010;
vector<int> ed[maxn];
int val[maxn];
int c[maxn];
bool cmp(int a, int b) { return val[a] < val[b]; }
vector<int> dfs(int v, int p) {
  vector<int> cur;
  for (int u : ed[v]) {
    if (u == p) {
      continue;
    }
    vector<int> a = dfs(u, v);
    for (int x : a) {
      cur.push_back(x);
    }
  }
  sort(cur.begin(), cur.end(), cmp);
  if (!cur.empty()) val[cur[0]] = 2;
  for (int i = 1; i < (int)cur.size(); i++) {
    int x = cur[i];
    int y = cur[i - 1];
    val[x] = val[y] + 2;
  }
  if (c[v] > (int)cur.size()) {
    cout << "NO\n";
    exit(0);
  }
  if (c[v] == 0) {
    val[v] = 1;
  } else {
    val[v] = val[cur[c[v] - 1]] + 1;
  }
  cur.push_back(v);
  return cur;
}
int main() {
  int n;
  cin >> n;
  int r = -1;
  for (int i = 1; i <= n; i++) {
    int pr;
    scanf("%d %d", &pr, &c[i]);
    if (pr != 0)
      ed[pr].push_back(i);
    else
      r = i;
  }
  dfs(r, -1);
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%d ", val[i]);
  }
  return 0;
}
