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;
vector<int> g[2010];
int n, c[2010], a[2010], sz[2010];
int l[2010], r[2010], t = 0;
bool ok = 1;
struct node {
  int v, id;
} b[2010];
bool cmp(const node& x, const node& y) { return x.v < y.v; }
void dfs(int rt) {
  if (ok == 0) return;
  ++t;
  l[rt] = t;
  b[t].id = rt;
  b[t].v = 0;
  int i, dd = 0;
  sz[rt]++;
  for (auto u : g[rt]) {
    dfs(u);
    sz[rt] += sz[u];
    for (i = l[u]; i <= r[u]; i++) {
      b[i].v += dd;
    }
    dd += sz[u];
  }
  r[rt] = t;
  if (c[rt] >= sz[rt]) {
    ok = 0;
    return;
  }
  sort(b + 1 + l[rt], b + 1 + r[rt], cmp);
  b[l[rt]].v = b[l[rt] + c[rt]].v + 1;
  for (i = l[rt] + c[rt] + 1; i <= r[rt]; i++) {
    b[i].v++;
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    int p;
    scanf("%d%d", &p, &c[i]);
    g[p].push_back(i);
  }
  int rt = g[0][0];
  dfs(rt);
  if (!ok)
    printf("NO\n");
  else {
    printf("YES\n");
    for (i = 1; i <= n; i++) a[b[i].id] = b[i].v;
    for (i = 1; i <= n; i++) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}
