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
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 = 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)))
