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 = 2010;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int n, root, cnt, fir[N], c[N], fa[N], sub[N], id[N][N], m[N], tmp[N], val[N];
struct edge {
  int to, nxt;
} e[N];
void add(int a, int b) {
  e[++cnt].to = b;
  e[cnt].nxt = fir[a];
  fir[a] = cnt;
}
void dfs(int u) {
  sub[u] = 1;
  for (register int i = fir[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    dfs(v);
    sub[u] += sub[v];
    for (register int j = 1; j <= m[v]; j++) id[u][++m[u]] = id[v][j];
  }
  if (c[u] >= sub[u]) {
    puts("NO");
    exit(0);
  }
  for (register int i = m[u]; i >= c[u] + 1; i--) id[u][i + 1] = id[u][i];
  id[u][c[u] + 1] = u;
  m[u]++;
}
int main() {
  n = read();
  for (register int i = 1; i <= n; i++) {
    fa[i] = read(), c[i] = read();
    if (!fa[i])
      root = i;
    else
      add(fa[i], i);
  }
  dfs(root);
  for (register int i = 1; i <= n; i++) val[id[root][i]] = i;
  puts("YES");
  for (register int i = 1; i <= n; i++) printf("%d ", val[i]);
  return 0;
}
