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
class ordered_set:

    def __init__(self, N):
        self.N = N+1
        self.tree = [0 for i in range(self.N)]

    def add(self, idx):
        while (idx < self.N):
            self.tree[idx] += 1
            idx += (idx & -idx)

    def remove(self, idx):
        while (idx < self.N):
            self.tree[idx] -= 1
            idx += (idx & -idx)

    def find_by_order(self, k):
        k -= 1
        idx = 0
        mask = 1

        while ((mask<<1) < self.N):
            mask <<= 1

        while (mask):
            ti = idx + mask
            mask >>= 1
            if (ti < self.N):
                if (k >= self.tree[ti]):
                    idx = ti
                    k -= self.tree[ti]

        if (k != 0):
            return -1
        return idx+1


def subtree_size(cur, adj, size):
    size[cur] = 1
    for v in adj[cur]:
        if (size[v] == 0):
            subtree_size(v, adj, size)
            size[cur] += size[v]


def fill(cur, adj, ost, c, a):
    a[cur] = ost.find_by_order(c[cur]+1)
    ost.remove(a[cur])
    for v in adj[cur]:
        if (a[v] == 0):
            fill(v, adj, ost, c, a)


n = int(input())
adj = [[] for i in range(n+1)]
c = [0]
for i in range(1,n+1):
    p, ct = map(int,input().split())
    adj[p].append(i)
    c.append(ct)

size = [0 for i in range(n+1)]
subtree_size(adj[0][0], adj, size)

ok = True
for i in range(1,n+1):
    if (c[i] >= size[i]):
        ok = False
        break

if not ok:
    print("NO")
else:
    ost = ordered_set(n)
    for i in range(1,n+1):
        ost.add(i)

    a = [0 for i in range(n+1)]
    fill(adj[0][0], adj, ost, c, a)

    print("YES")
    for i in range(1,n+1):
        print(a[i], end=' ')
    print()
