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 double eps = 1e-10;
const int maxn = 2000 + 10;
const long long mod = 1e9 + 7;
int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }
using namespace std;
struct edge {
  int v, nxt;
} e[maxn << 1];
int cnt;
int head[maxn];
int c[maxn], siz[maxn], p[maxn];
int vis[maxn];
int tot, n;
inline void add_edge(int u, int v) {
  e[++cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
inline void get_siz(int u) {
  siz[u] = 1;
  for (int i = head[u]; ~i; i = e[i].nxt) {
    int v = e[i].v;
    get_siz(v);
    siz[u] += siz[v];
  }
}
inline void dfs(int u) {
  if (c[u] > siz[u] - 1) {
    cout << "NO" << endl;
    exit(0);
  }
  int xx = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) xx++;
    if (xx == c[u] + 1) {
      p[u] = i;
      vis[i] = 1;
      break;
    }
  }
  for (int i = head[u]; ~i; i = e[i].nxt) dfs(e[i].v);
}
int main() {
  scanf("%d", &n);
  int root;
  memset(head, -1, sizeof(head));
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    int f;
    scanf("%d%d", &f, &c[i]);
    if (f == 0)
      root = i;
    else
      add_edge(f, i);
  }
  get_siz(root);
  tot = 1;
  dfs(root);
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", p[i]);
  printf("\n");
  return 0;
}
