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 double pi = acos(-1.0);
const int mod = 998244353;
void reduce(int &a) { a += a >> 31 & mod; }
long long Mod(long long x) { return x >= mod ? x - mod : x; }
long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return x * f;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 2010, M = N;
int n, c[N], root;
;
struct edge {
  int nxt, to, w;
} e[M << 1];
int head[N], ecnt = 1;
void add(int u, int v, long long w = 0) {
  e[++ecnt].w = w, e[ecnt].to = v, e[ecnt].nxt = head[u];
  head[u] = ecnt;
}
void add_edge(int u, int v, long long w = 0) {
  add(u, v, w), add(v, u, w * 1);
};
int siz[N], a[N];
void no_ans() {
  puts("NO");
  exit(0);
}
vector<int> dfs(int u) {
  siz[u] = 1;
  vector<int> ve;
  int num = 0;
  if (!c[u]) ve.push_back(u);
  for (int i = head[u], v; i; i = e[i].nxt) {
    vector<int> h = dfs(v = e[i].to);
    siz[u] += siz[v];
    for (int j : h) {
      num++, ve.push_back(j);
      if (num == c[u]) ve.push_back(u);
    }
  }
  if (c[u] > siz[u] - 1) no_ans();
  return ve;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int p = read();
    c[i] = read();
    if (!p)
      root = i;
    else
      add(p, i);
  }
  vector<int> ans = dfs(root);
  int num = 0;
  puts("Yes");
  for (int i : ans) a[i] = ++num;
  for (int i = 1; i <= n; i++) print(a[i]), putchar(' ');
  return 0;
}
