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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
struct node {
  int t, nt;
} e[4005 << 1];
int n, head[4005], tot, c[4005], sz[4005], ans[4005];
inline void add(int u, int v) {
  e[++tot].t = v;
  e[tot].nt = head[u];
  head[u] = tot;
}
inline void dfs2(int u, int fa, int w) {
  if (ans[u] >= w) ans[u]++;
  for (int i = head[u]; i; i = e[i].nt) {
    int v = e[i].t;
    if (v == fa) continue;
    dfs2(v, u, w);
  }
}
inline void dfs3(int u, int fa, int w) {
  ans[u] += w;
  for (int i = head[u]; i; i = e[i].nt) {
    int v = e[i].t;
    if (v == fa) continue;
    dfs3(v, u, w);
  }
}
inline void dfs(int u, int fa) {
  sz[u] = 1;
  for (int i = head[u]; i; i = e[i].nt) {
    int v = e[i].t;
    if (v == fa) continue;
    dfs(v, u);
    dfs3(v, u, sz[u] - 1);
    sz[u] += sz[v];
  }
  if (c[u] >= sz[u]) puts("NO"), exit(0);
  dfs2(u, fa, c[u] + 1);
  ans[u] = c[u] + 1;
}
int rt;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int fa = read();
    c[i] = read();
    if (fa)
      add(fa, i), add(i, fa);
    else
      rt = i;
  }
  dfs(rt, 0);
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
