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 long long P = 998244353;
const int N = 2005;
mt19937 rng(time(0));
vector<int> g[N];
int a[N], sz[N], ans[N];
int n, x, y;
bitset<N> use, bkp[N];
void dfs(int x) {
  sz[x] = 1;
  for (auto u : g[x]) {
    dfs(u);
    sz[x] += sz[u];
  }
}
void dfs2(int x) {
  int cnt = 0;
  for (int i = (1); i <= (n); ++i) {
    cnt += use[i];
    if (cnt == a[x] + 1) {
      ans[x] = i;
      use[i] = 0;
      break;
    }
  }
  bkp[x] = use;
  use = 0;
  int now = 1, last;
  for (auto u : g[x]) {
    cnt = 0;
    last = now;
    while (cnt < sz[u]) {
      cnt += bkp[x][now];
      now++;
    }
    for (int i = (last); i <= (now); ++i) use[i] = bkp[x][i];
    dfs2(u);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d%d", &x, &y);
    g[x].push_back(i);
    a[i] = y;
  }
  dfs(g[0][0]);
  for (int i = (1); i <= (n); ++i) {
    if (a[i] >= sz[i]) {
      printf("NO");
      return 0;
    }
  }
  printf("YES\n");
  use.set();
  dfs2(g[0][0]);
  for (int i = (1); i <= (n); ++i) printf("%d ", ans[i]);
  return 0;
}
