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
n = int(input())
parent = [0] * (n + 1)
great = [0] * (n + 1)
tree = [[] for _ in range(n + 1)]
root = 0
for i in range(1, n + 1):
    p, c = map(int, input().split())
    tree[p].append(i)
    if p == 0:
        root = i
    parent[i] = p
    great[i] = c
stack = [root]
ptr = [0]
accum = [[], []]
while len(stack) > 0:
    v = stack[-1]
    if ptr[-1] == len(tree[v]):
        if great[v] > len(accum[-1]):
            print("NO")
            exit()
        accum[-1].insert(great[v], v)
        accum[-2] += accum[-1]
        accum.pop()
        stack.pop()
        ptr.pop()
        continue
    to = tree[v][ptr[-1]]
    ptr[-1] += 1
    stack.append(to)
    ptr.append(0)
    accum.append([])
print("YES")
ans = [None] * n
for i in range(n):
    ans[accum[0][i] - 1] = str(i + 1)
print(' '.join(ans))
