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>
template <class T>
inline void read(T &res) {
  res = 0;
  bool bo = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    bo = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  if (bo) res = ~res + 1;
}
const int N = 2005;
int n, ecnt, nxt[N], adj[N], go[N], sze[N], c[N], rt, a[N];
bool flag;
void add_edge(int u, int v) {
  nxt[++ecnt] = adj[u];
  adj[u] = ecnt;
  go[ecnt] = v;
}
void calcsze(int u) {
  sze[u] = 1;
  for (int e = adj[u], v = go[e]; e; e = nxt[e], v = go[e])
    calcsze(v), sze[u] += sze[v];
  if (c[u] >= sze[u]) flag = 1;
}
void dfs(int u, std::vector<int> val) {
  a[u] = val[c[u]];
  int tx = 0;
  for (int e = adj[u], v = go[e]; e; e = nxt[e], v = go[e]) {
    std::vector<int> nxt;
    for (int i = 0; i < sze[v]; i++) {
      if (tx == c[u]) tx++;
      nxt.push_back(val[tx]);
      tx++;
    }
    dfs(v, nxt);
  }
}
int main() {
  int x;
  read(n);
  for (int i = 1; i <= n; i++) {
    read(x);
    read(c[i]);
    if (x)
      add_edge(x, i);
    else
      rt = i;
  }
  calcsze(rt);
  if (flag) return puts("NO"), 0;
  std::vector<int> init;
  for (int i = 1; i <= n; i++) init.push_back(i);
  dfs(rt, init);
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return puts(""), 0;
}
