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;
mt19937 Rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename T>
inline void chkmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
inline void chkmin(T &x, T y) {
  if (x > y) x = y;
}
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < 48) c = getchar();
  while (c > 47) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
const int maxn = 2010;
int n, rt, fa[maxn], sz[maxn], cnt[maxn];
vector<int> e[maxn], vec[maxn];
void dfs(int u) {
  sz[u] = 1;
  for (int v : e[u]) {
    dfs(v);
    sz[u] += sz[v];
    for (int x : vec[v]) vec[u].push_back(x);
  }
  if (cnt[u] >= sz[u]) {
    puts("NO");
    exit(0);
  }
  vec[u].insert(vec[u].begin() + cnt[u], u);
}
int main() {
  n = read();
  for (int i = (1), iend = (n); i <= iend; ++i) {
    fa[i] = read(), cnt[i] = read();
    if (fa[i]) {
      e[fa[i]].push_back(i);
    } else {
      rt = i;
    }
  }
  dfs(rt);
  puts("YES");
  int cur = 0;
  static int ans[maxn];
  for (int x : vec[rt]) {
    ans[x] = ++cur;
  }
  for (int i = (1), iend = (n); i <= iend; ++i) printf("%d ", ans[i]);
  return 0;
}
