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;
bool funfa = true;
int prox[2010];
int pai[2010];
int c[2010];
vector<int> filhos[2010];
int desc[2010];
pair<int, int> constroi(int v) {
  if (filhos[v].size() == 0) return make_pair(v, v);
  int primeiro, ultimo;
  for (int i = 0; i < filhos[v].size(); i++) {
    pair<int, int> u = constroi(filhos[v][i]);
    if (i == 0) {
      primeiro = u.first;
      ultimo = u.second;
    } else {
      prox[ultimo] = u.first;
      ultimo = u.second;
    }
  }
  int pos = primeiro;
  if (c[v] == 0 || c[v] == desc[v] - 1) {
    if (c[v] == 0) {
      prox[v] = primeiro;
      primeiro = v;
    } else {
      prox[ultimo] = v;
      ultimo = v;
    }
  } else {
    for (int i = 1; i < c[v]; i++) pos = prox[pos];
    prox[v] = prox[pos];
    prox[pos] = v;
  }
  return pair<int, int>(primeiro, ultimo);
}
void dfs(int v) {
  int tot = 1;
  for (int i = 0; i < filhos[v].size(); i++) {
    dfs(filhos[v][i]);
    tot += desc[filhos[v][i]];
  }
  desc[v] = tot;
  if (c[v] >= desc[v]) funfa = false;
}
int main() {
  int n;
  scanf("%d", &n);
  int raiz;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &pai[i], &c[i]);
    filhos[pai[i]].push_back(i);
    if (pai[i] == 0) raiz = i;
  }
  dfs(raiz);
  if (!funfa)
    printf("NO\n");
  else {
    printf("YES\n");
    int u = constroi(raiz).first;
    int id[2020];
    for (int i = 1; i <= n; i++) {
      id[u] = i;
      u = prox[u];
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", id[i]);
    }
    printf("\n");
  }
}
