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 deque
N = int(input())
X = [[] for i in range(N)]
P = [-1] * N
C = [0] * N
for i in range(N):
    p, c = map(int, input().split())
    C[i] = c
    if p:
        p -= 1
        P[i] = p
        X[p].append(i)
    else:
        rt = i

Q = deque([rt])
R = []
while Q:
    i = deque.popleft(Q)
    R.append(i)
    for a in X[i]:
        P[a] = i
        deque.append(Q, a)

def calc(i, cc, AA):
    B = {}
    k = 0
    for A in AA:
        for a in A: B[a] = A[a]+k
        k += len(A)
    if cc > len(B):
        print("NO")
        exit()
    B = {a: (B[a] if B[a] < cc else B[a] + 1) for a in B}
    B[i] = cc
    return B

L = [[] for _ in range(N)]
for i in R[::-1]:
    K = []
    for j in X[i]:
        K.append(L[j])
    L[i] = calc(i, C[i], K)

print("YES")
ANS = [0] * N
for a in L[rt]:
    ANS[a] = L[rt][a] + 1

print(*ANS)