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
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import defaultdict,deque
def bfs(node):
    vis[node]=1
    q=deque([node])
    order.append(node)
    while q:
        cur=q.popleft()
        for j in edge[cur]:
            if vis[j]==0:
                child[cur].append(j)
                vis[j]=1
                order.append(j)
                q.append(j)
def dfs(node):
    vis[node]=1
    stack=[node]
    while stack:
        cur=stack.pop()
        ans[cur-1]=lis[arr[cur]]
        lis.remove(ans[cur-1])
        for j in edge[cur]:
            if vis[j]==0:
                vis[j]=1
                stack.append(j)
n=int(input())
edge=defaultdict(list)
child=defaultdict(list)
root=-1
arr=[-1]*(n+1)
for i in range(n):
    p,c=list(map(int,input().split()))
    arr[i+1]=c
    if p!=0:
        edge[p].append(i+1)
        edge[i+1].append(p)
    else:
        root=i+1
subTreeSize=[1]*(n+1)
vis=[0]*(n+1)
order=[]
bfs(root)
order.reverse()
for i in range(n):
    for j in child[order[i]]:
        subTreeSize[order[i]]+=subTreeSize[j]
x=0
for i in range(1,n+1):
    if arr[i]>subTreeSize[i]-1:
        x+=1
        break
if x==1:
    print("NO")
else:
    print("YES")
    lis=[]
    for i in range(n):
        lis.append(i+1)
    vis=[0]*(n+1)
    ans=[-1]*n
    dfs(root)
    print(" ".join(str(x) for x in ans))