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 MOD = 1000000007;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int iNF = 0x3f3f3f3f;
const long long MAXN = 2003;
int n, p, c[MAXN], ans[MAXN], sz[MAXN];
vector<int> edge[MAXN];
bool hasAns = true;
void build_sz(int nd) {
  sz[nd] = 1;
  for (auto v : edge[nd]) {
    build_sz(v);
    sz[nd] += sz[v];
  }
}
void dfs(int nd, vector<int> rem) {
  ;
  if (c[nd] >= int(rem.size())) {
    hasAns = false;
  } else {
    vector<int> nw;
    for (int i = 0; i < (int(rem.size())); i += 1) {
      if (i != c[nd]) {
        nw.emplace_back(rem[i]);
      }
    }
    ans[nd] = rem[c[nd]];
    int idx = 0;
    for (auto v : edge[nd]) {
      vector<int> cur;
      for (int j = 0; j < (sz[v]); j += 1) {
        cur.emplace_back(nw[idx]);
        idx++;
      }
      dfs(v, cur);
    }
  }
}
int main() {
  ;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int root = -1;
  for (int i = 1; i < (n + 1); i += 1) {
    cin >> p >> c[i];
    if (p) {
      edge[p].emplace_back(i);
    } else {
      root = i;
    }
  }
  vector<int> rem;
  for (int i = 1; i < (n + 1); i += 1) {
    rem.emplace_back(i);
  }
  build_sz(root);
  dfs(root, rem);
  if (hasAns) {
    cout << "YES" << '\n';
    for (int i = 1; i < (n + 1); i += 1) {
      cout << ans[i] << " \n"[i == n];
    }
  } else {
    cout << "NO" << '\n';
  }
  return 0;
}
