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>
const int Maxn = 2000;
int sz[Maxn + 5];
int fa[Maxn + 5];
int head[Maxn + 5], arrive[Maxn + 5], e_nxt[Maxn + 5], tot;
void add_edge(int from, int to) {
  arrive[++tot] = to;
  e_nxt[tot] = head[from];
  head[from] = tot;
}
int n, Root;
int c[Maxn + 5];
int nxt[Maxn + 5], pre[Maxn + 5];
int lea[Maxn + 5], lea_len;
int a[Maxn + 5];
int dfn[Maxn + 5], out[Maxn + 5], dfn_tot;
void init_dfs(int u) {
  sz[u] = 1;
  for (int i = head[u]; i; i = e_nxt[i]) {
    int v = arrive[i];
    init_dfs(v);
    sz[u] += sz[v];
  }
  if (sz[u] == 1) {
    lea[++lea_len] = u;
  }
}
void dfs(int u) {
  dfn[u] = ++dfn_tot;
  for (int i = head[u]; i; i = e_nxt[i]) {
    int v = arrive[i];
    dfs(v);
  }
  out[u] = dfn_tot;
  if (sz[u] == 1) {
    return;
  }
  int pos = 0;
  int num = 0;
  while (num < c[u] || (num == c[u] && c[u] != sz[u] - 1)) {
    pos = nxt[pos];
    if (dfn[u] <= dfn[pos] && out[u] >= dfn[pos]) {
      num++;
    }
  }
  if (c[u] == sz[u] - 1) {
    pos = nxt[pos];
  }
  int las = pre[pos];
  nxt[u] = pos;
  pre[u] = las;
  nxt[las] = u;
  pre[pos] = u;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &fa[i], &c[i]);
    if (fa[i] == 0) {
      Root = i;
    }
    add_edge(fa[i], i);
  }
  init_dfs(Root);
  lea[lea_len + 1] = n + 1;
  for (int i = 1; i <= lea_len; i++) {
    pre[lea[i]] = lea[i - 1];
    nxt[lea[i]] = lea[i + 1];
  }
  nxt[0] = lea[1];
  pre[n + 1] = lea[lea_len];
  for (int i = 1; i <= n; i++) {
    if (c[i] >= sz[i]) {
      puts("NO");
      return 0;
    }
  }
  dfs(Root);
  int id_tot = 0;
  int pos = nxt[0];
  while (pos != n + 1) {
    a[pos] = ++id_tot;
    pos = nxt[pos];
  }
  puts("YES");
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[i]);
  }
  puts("");
  return 0;
}
