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;
template <typename T>
bool get_int(T &x) {
  char t = getchar();
  bool neg = false;
  x = 0;
  for (; (t > '9' || t < '0') && t != '-' && t != EOF; t = getchar())
    ;
  if (t == '-') neg = true, t = getchar();
  if (t == EOF) return false;
  for (; t <= '9' && t >= '0'; t = getchar()) x = x * 10 + t - '0';
  if (neg) x = -x;
  return true;
}
template <typename T>
void print_int(T x) {
  if (x < 0) putchar('-'), x = -x;
  short a[20] = {}, sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
const int inf = 0x3f3f3f3f;
const long long Linf = 1ll << 61;
const double pi = acos(-1.0);
const int maxn = 2111;
vector<int> g[maxn];
int n, p[maxn], c[maxn], sz[maxn], ans[maxn];
void dfs(int x) {
  sz[x] = 1;
  for (auto v : g[x]) {
    dfs(v);
    sz[x] += sz[v];
  }
  if (c[x] >= sz[x]) {
    puts("NO");
    exit(0);
  }
}
void dfs2(int x, vector<int> &vs) {
  ans[x] = vs[c[x]];
  int cur = 0;
  for (auto v : g[x]) {
    vector<int> nv;
    for (int i = 0; i < sz[v]; i++) {
      if (vs[cur] == ans[x]) cur++;
      nv.push_back(vs[cur]);
      cur++;
    }
    dfs2(v, nv);
  }
}
int main() {
  get_int(n);
  int rt = 0;
  for (int i = 1; i <= n; i++) {
    (get_int(p[i]) && get_int(c[i]));
    if (p[i])
      g[p[i]].push_back(i);
    else
      rt = i;
  }
  dfs(rt);
  vector<int> vs;
  for (int i = 1; i <= n; i++) vs.push_back(i);
  dfs2(rt, vs);
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
