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 long long mod = 998244353;
const int maxn = 2e3 + 5;
int vis[maxn];
struct Edge {
  int v;
  int next;
} edge[maxn];
int num;
int head[maxn];
int tol;
int val[maxn];
int cc[maxn];
int flag;
void addEdge(int u, int v) {
  edge[tol].v = v;
  edge[tol].next = head[u];
  head[u] = tol++;
}
int cnt[maxn];
int fun(int x) {
  int cnt1 = 0;
  for (int i = 1; i <= num; i++) {
    if (vis[i] == 0) cnt1++;
    if (cnt1 == x) {
      cnt1 = i;
      break;
    }
  }
  return cnt1;
}
void dfs1(int x) {
  cnt[x] = 1;
  for (int i = head[x]; i != 0; i = edge[i].next) {
    int v = edge[i].v;
    dfs1(v);
    cnt[x] += cnt[v];
  }
}
void dfs2(int x) {
  if (cnt[x] - 1 < cc[x]) {
    printf("NO\n");
    exit(0);
  }
  int ele = fun(cc[x] + 1);
  vis[ele] = 1;
  val[x] = ele;
  for (int i = head[x]; i != 0; i = edge[i].next) {
    int v = edge[i].v;
    dfs2(v);
  }
}
int main() {
  scanf("%d", &num);
  int root;
  flag = 0;
  tol = 1;
  for (int i = 1; i <= num; i++) {
    int u, c;
    scanf("%d%d", &u, &c);
    if (u == 0) {
      root = i;
    } else
      addEdge(u, i);
    cc[i] = c;
  }
  dfs1(root);
  dfs2(root);
  printf("YES\n");
  for (int i = 1; i <= num; i++) printf("%d ", val[i]);
  printf("\n");
}
