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
from collections import  defaultdict
def dfs(g,node,parent,trees):
	l=0
	for i in g[node]:
		if i!=parent:
			l+=dfs(g,i,node,trees)
	trees[node]=l
	return l+1
def dfs2(g,node,Small,answer,C,trees):
	answer[node]=Small[C[node]]
	supersmall=[]
	for i in range(len(Small)):
		if i!=C[node]:
			supersmall.append(Small[i])
	cnts=0
	for i in g[node]:
		u=[]
		for j in range(trees[i]):
			u.append(supersmall[cnts])
			cnts+=1
		dfs2(g,i,u,answer,C,trees)


n=int(input())
root=None
C=[0]*n
g=defaultdict(list)
for i in range(n):
	parent,c=map(int,input().strip().split())
	if parent==0:
		root=i
	else:
		g[parent-1].append(i)
	C[i]=c

trees=[0]*(n)
dfs(g,root,-1,trees)
check=list(map(lambda x,y:x-y,trees,C))
for i in check:
	if i<0:
		print("NO")
		exit(0)
ans=[0]*n
small=[* range(1,n+1)]
trees=list(map(lambda s:s+1,trees))
dfs2(g,root,small,ans,C,trees)
print("YES")
print(*ans)