human
Can you solve the below in PYTHON3?
Frog Gorf is traveling through Swamp kingdom. Unfortunately, after a poor jump, he fell into a well of n meters depth. Now Gorf is on the bottom of the well and has a long way up.

The surface of the well's walls vary in quality: somewhere they are slippery, but somewhere have convenient ledges. In other words, if Gorf is on x meters below ground level, then in one jump he can go up on any integer distance from 0 to a_x meters inclusive. (Note that Gorf can't jump down, only up).

Unfortunately, Gorf has to take a break after each jump (including jump on 0 meters). And after jumping up to position x meters below ground level, he'll slip exactly b_x meters down while resting.

Calculate the minimum number of jumps Gorf needs to reach ground level.

Input

The first line contains a single integer n (1 ≤ n ≤ 300 000) — the depth of the well.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i), where a_i is the maximum height Gorf can jump from i meters below ground level.

The third line contains n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ n - i), where b_i is the distance Gorf will slip down if he takes a break on i meters below ground level.

Output

If Gorf can't reach ground level, print -1. Otherwise, firstly print integer k — the minimum possible number of jumps.

Then print the sequence d_1,\,d_2, …,\,d_k where d_j is the depth Gorf'll reach after the j-th jump, but before he'll slip down during the break. Ground level is equal to 0.

If there are multiple answers, print any of them.

Examples

Input

3
0 2 2
1 1 0

Output

2
1 0 

Input

2
1 1
1 0

Output

-1

Input

10
0 1 2 3 5 5 6 7 8 5
9 8 7 1 5 4 3 2 0 0

Output

3
9 4 0 

Note

In the first example, Gorf is on the bottom of the well and jump to the height 1 meter below ground level. After that he slip down by meter and stays on height 2 meters below ground level. Now, from here, he can reach ground level in one jump.

In the second example, Gorf can jump to one meter below ground level, but will slip down back to the bottom of...
 Give me a Python code to print -1?
assistant
#Here is Python code to print -1.

import sys
input = lambda : sys.stdin.readline().rstrip()

write = lambda x: sys.stdout.write(x+"\n")
debug = lambda x: sys.stderr.write(x+"\n")
writef = lambda x: print("{:.12f}".format(x))
# sys.setrecursionlimit(3*10**5+10)
### セグメント木(はやい)
class SG:
    def __init__(self, n, v=None):
        self._n = n
        self.geta = 0
        x = 0
        while (1 << x) < n:
            x += 1
        self._log = x
        self._size = 1 << self._log
        self._d = [ninf] * (2 * self._size)
        if v is not None:
            for i in range(self._n):
                self._d[self._size + i] = v[i]
        for i in range(self._size - 1, 0, -1):
            self._update(i)
    def _update(self, k):
        self._d[k] = op(self._d[2 * k], self._d[2 * k + 1])
    def update(self, p, x):
        assert 0 <= p < self._n
#         x -= self.geta
        p += self._size
        self._d[p] = x
        for i in range(1, self._log + 1):
#             self._update(p >> i)
            k = p>>i
            self._d[k] = op(self._d[2 * k], self._d[2 * k + 1])
    def get(self, p):
        assert 0 <= p < self._n
        return self._d[p + self._size] # + self.geta
    def check(self):
        return [self.get(p) for p in range(self._n)]
    def query(self, left, right):
        # [l,r)の総和
        assert 0 <= left <= right <= self._n
        sml = ninf
        smr = ninf
        left += self._size
        right += self._size
        # 外側から計算していく(lは小さい側から, rは大きい側から)
        while left < right:
            if left & 1:
                sml = op(sml, self._d[left])
                left += 1
            if right & 1:
                right -= 1
                smr = op(self._d[right], smr)
            left >>= 1
            right >>= 1
        return op(sml, smr) # + self.geta
#     def update_all(self, v):
#         # 全体加算
#         self.geta += v
    def query_all(self):
        return self._d[1] # + self.geta
    def max_right(self, left, f):
        """f(op(a[l], a[l + 1], ..., a[r - 1])) = true となる最大の r
        ->...