human
Give me a Python code to print "YES", and in the second line output n integers a_i (1 ≤ a_i ≤ {10}^{9}) given the following:
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. 

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
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python code to print "YES", and in the second line output n integers a_i (1 ≤ a_i ≤ {10}^{9}).

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

def main():
    
    n=int(input())
    p=[-1 for _ in range(n+1)]
    c=[-1 for _ in range(n+1)]
    children=[[] for _ in range(n+1)]
    for i in range(1,n+1):
        pp,cc=readIntArr()
        p[i]=pp
        c[i]=cc
        if pp==0: #root
            root=i
        else:
            children[pp].append(i)
    
    @bootstrap
    def dfs(i): # return [subtree nodes]
        # print('node:{}'.format(i))
        if len(children[i])==0: # leaf, no children
            if c[i]>0:
                possible[0]=False
            val=1
            # "push" all larger values than val by 1
            for zz in range(1,n+1):
                if a[zz]>=val:
                    a[zz]+=1
            a[i]=val
            # print('leaf:{} a:{}'.format(i,a))
            yield [i]
        else:
            requiredCnts=c[i]
            childNodes=[]
            for child in children[i]:
                for node in (yield dfs(child)):
                    childNodes.append(node)
            if requiredCnts>len(childNodes):
                possible[0]=False
                yield [] # impossible
            childVals=[a[node] for node in childNodes]
            childVals.sort()
            if requiredCnts==0:
                val=1
            else:
                val=childVals[requiredCnts-1]+1
            # "push" all larger values than val by 1
            for zz in range(1,n+1):
                if a[zz]>=val:
                    a[zz]+=1
            
            #...