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 = (int)2002 + 2;
const int LG = 20;
const int mod = (int)1e9 + 7;
const int INF = (int)1e9 + 7;
const long long linf = 1e16;
const long double pi = acos(-1);
const int base = 10;
const long double EPS = 1e-9;
int n, c[N], sz[N], ans[N];
vector<int> g[N];
set<int> st;
void dfs_sz(int v, int p = -1) {
  sz[v] = 1;
  for (int to : g[v]) {
    if (to == p) continue;
    dfs_sz(to, v);
    sz[v] += sz[to];
  }
}
void dfs(int v, int p = -1) {
  int pos = 0;
  for (auto to : st) {
    if (pos == c[v]) {
      ans[v] = to;
      break;
    } else
      ++pos;
  }
  st.erase(ans[v]);
  for (int to : g[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
}
int main() {
  scanf("%d", &n);
  int root = -1;
  for (int i = 1; i <= n; ++i) {
    int p;
    scanf("%d %d", &p, c + i);
    if (p == 0) {
      root = i;
      continue;
    }
    g[p].emplace_back(i);
    g[i].emplace_back(p);
  }
  dfs_sz(root);
  for (int i = 1; i <= n; ++i)
    if (sz[i] - 1 < c[i]) return !printf("NO\n");
  printf("YES\n");
  for (int i = 1; i <= n; ++i) st.insert(i);
  dfs(root);
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
