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;
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    que;
stringstream ssin;
const long long LINF = 0x7fffffffffffffffll;
const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int n, m, rt, idx, flag;
int c[N];
int e[M], ne[M], h[N], sz[N];
int temp[N], ans[N];
inline long long read() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int find(int x) {
  int t = 0;
  for (int i = 1; i <= n; ++i) {
    if (temp[i] == 0) continue;
    t++;
    if (t == x) {
      temp[i] = 0;
      return i;
    }
  }
  return -1;
}
void dfs(int u) {
  sz[u] = 1;
  ans[u] = find(c[u] + 1);
  if (ans[u] == -1) flag = -1;
  for (int i = h[u]; ~i; i = ne[i]) {
    int v = e[i];
    dfs(v);
    sz[u] += sz[v];
  }
  if (sz[u] - 1 < c[u]) flag = -1;
}
void add(int a, int b) {
  e[idx] = b;
  ne[idx] = h[a];
  h[a] = idx++;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) temp[i] = i;
  memset(h, -1, sizeof h);
  ;
  for (int i = 1; i <= n; ++i) {
    int p;
    p = read();
    c[i] = read();
    if (!p) {
      rt = i;
      continue;
    }
    add(p, i);
  }
  dfs(rt);
  if (flag == -1)
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
    puts("");
  }
}
