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 $n$ $(1 \leq n \leq 2000)$ — 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 $c_i$ ($0 \leq p_i \leq n$; $0 \leq c_i \leq n-1$), 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 $a_i$ $(1 \leq a_i \leq {10}^{9})$. 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

Difficulty: competition

Starter code:

Solutions:
import sys
readline = sys.stdin.readline        
def parorder(Edge, p):
    N = len(Edge)
    par = [0]*N
    par[p] = -1
    stack = [p]
    order = []
    visited = set([p])
    ast = stack.append
    apo = order.append
    while stack:
        vn = stack.pop()
        apo(vn)
        for vf in Edge[vn]:
            if vf in visited:
                continue
            visited.add(vf)
            par[vf] = vn
            ast(vf)
    return par, order

def getcld(p):
    res = [[] for _ in range(len(p))]
    for i, v in enumerate(p[1:], 1):
        res[v].append(i)
    return res



N = int(readline())
root = None
Edge = [[] for _ in range(N)]
Cr = [None]*N
for a in range(N):
    b, c = list(map(int, readline().split()))
    b -= 1
    if b == -1:
        root = a
    else:
        Edge[a].append(b)
        Edge[b].append(a)
    Cr[a] = c

P, L = parorder(Edge, root)
        
dp = [0]*N

for l in L[:0:-1]:
    p = P[l]
    dp[p] += 1+dp[l]

if any(d < c for d, c in zip(dp, Cr)):
    print('NO')
else:
    print('YES')
    A = [None]*N
    dp2 = [[] for _ in range(N)]
    for l in L[:0:-1]:
        p = P[l]
        dp2[l] = dp2[l][:Cr[l]] + [l] + dp2[l][Cr[l]:]
        dp2[p].extend(dp2[l])
    dp2[root] = dp2[root][:Cr[root]] + [root] + dp2[root][Cr[root]:]        
    Ans = [None]*N
    for i in range(N):
        Ans[dp2[root][i]] = i+1
    print(' '.join(map(str, Ans)))