Nauuo and Cards

传送门

Nauuo is a girl who loves playing cards.

One day she was playing cards but found that the cards were mixed with some empty ones.

There are $n$ cards numbered from $1$ to $n$, and they were mixed with another $n$ empty cards. She piled up the $2n$ cards and drew $n$ of them. The $n$ cards in Nauuo’s hands are given. The remaining $n$ cards in the pile are also given in the order from top to bottom.

In one operation she can choose a card in her hands and play it — put it at the bottom of the pile, then draw the top card from the pile.

Nauuo wants to make the $n$ numbered cards piled up in increasing order (the $i$-th card in the pile from top to bottom is the card $i$) as quickly as possible. Can you tell her the minimum number of operations?

Input

The first line contains a single integer $n$ ($1\le n\le 2\cdot 10^5$) — the number of numbered cards.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($0\le a_i\le n$) — the initial cards in Nauuo’s hands. $0$ represents an empty card.

The third line contains $n$ integers $b_1,b_2,\dots ,b_n$ ($0\le b_i\le n$) — the initial cards in the pile, given in order from top to bottom. $0$ represents an empty card.

It is guaranteed that each number from $1$ to $n$ appears exactly once, either in $a_{\mathrm{1..}n}$ or $b_{\mathrm{1..}n}$.

Output

The output contains a single integer — the minimum number of operations to make the $n$ numbered cards piled up in increasing order.

Examples

input #1

3
0 2 0
3 0 1

output #1

2

input #2

3
0 2 0
1 0 3

output #2

4

input #3

11
0 0 0 5 0 0 0 4 0 0 11
9 2 6 0 8 1 7 0 3 0 10

output #3

18

Note

Example 1

We can play the card $2$ and draw the card $3$ in the first operation. After that, we have $[0,3,0]$ in hands and the cards in the pile are $[0,1,2]$ from top to bottom.

Then, we play the card $3$ in the second operation. The cards in the pile are $[1,2,3]$, in which the cards are piled up in increasing order.

Example 2

Play an empty card and draw the card $1$, then play $1$, $2$, $3$ in order.

题目翻译

Nauuo 是一个喜欢玩牌的女孩。

有一天，她正在玩牌，却发现牌里夹杂着一些空牌。

从 $1$ 到 $n$ 一共有 $n$ 张牌，它们和另外 $n$ 张空牌混在一起。她把 $2n$ 张牌堆在一起，抽出了其中的 $n$ 张。现给出诺诺手中的 $n$ 张牌。牌堆中剩下的 $n$ 张牌也按从上到下的顺序给出。

在一次操作中，她可以选择手中的一张牌并将其打出–将其放在牌堆的底部，然后从牌堆中抽取最上面的一张牌。

Nauuo 想尽快让 $n$ 张数字牌按递增顺序堆起来（从上到下牌堆中的第 $i$ 张牌是 $i$ 张牌）。你能告诉她最少的操作次数吗？

输入格式

第一行包含一个整数 $n$ ( $1\le n\le 2\cdot 10^5$ ) - 编号卡片的数量。

第二行包含 $n$ 个整数 $a_1,a_2,\dots ,a_n$ （ $0\le a_i\le n$ ）–诺诺手中的初始牌。 $0$ 代表空牌。

第三行包含 $n$ 个整数 $b_1,b_2,\dots ,b_n$ 。( $0\le b_i\le n$ ) -牌堆中的初始牌，从上到下依次排列。 $0$ 代表一张空牌。

可以保证从 $1$ 到 $n$ 的每个数字都正好出现一次，要么出现在 $a_{\mathrm{1..}n}$ 中，要么出现在 $b_{\mathrm{1..}n}$ 中。

输出格式

输出包含一个整数，即按递增顺序堆叠 $n$ 号纸牌的最少操作次数。

提示

样例 1

在第一次操作中，我们可以出牌 $2$ ，抽牌 $3$ 。之后，我们手中有 $[0,3,0]$ ，牌堆中的牌从上到下为 $[0,1,2]$ 。

然后，我们在第二次操作中出牌 $3$ 。牌堆中的牌为 $[1,2,3]$ ，其中牌按递增顺序堆叠。

样例 2

先出一张空牌，然后抽出牌 $1$ ，接着依次出 $1$ 、 $2$ 、 $3$ 。

注明

$以上来自CodeForces，翻译：DeepL、本人。$

解题思路

首先，在输入时记录牌堆中的每张牌 $b_i$ 初始在牌堆中的位置为 $P_{b_i}$，即将 $P_{b_i}$ 赋值为 $i$，看以下代码片段比较直观。

	for (int i = 1; i <= n; i++) cin >> b[i], P[b[i]] = i;

考虑是否使用空牌：

• 如果必须使用空牌，那么最好的办法就是把所有数字牌拿到手，再按顺序摆回牌堆，这样的操作次数是 $\max (P_i+1+n-i)$；
• 否则，次数为 $P_i-i$（$P_i$ 为 $P$ 数组中第一个满足条件： $P_i\ge i$ 的元素）。

AC Code

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int Maxn = 200000 + 5;
int n, a[Maxn], b[Maxn];
int P[Maxn];
int Answer_1, Answer_2;
inline void Init();
inline void Solve();
inline void Work() {Init(), Solve();cout << (Answer_2 == INT_MAX ? Answer_1 : Answer_2) << endl;
}
signed main() {FAST_IO,Work();return 0;
}
inline void Init() {cin >> n;for (int i = 1; i <= n; i++) cin >> a[i];for (int i = 1; i <= n; i++) cin >> b[i], P[b[i]] = i;Answer_1 = INT_MIN, Answer_2 = INT_MAX;
}
inline void Solve() {for (int i = 1, t; i <= n; i++) {t = P[i] + 1 + n - i;if (Answer_1 < t)Answer_1 = t;}for (int i = 1; i <= n; i++) if (P[i] >= i) {Answer_2 = P[i] - i;break;}for (int i = 1; i <= n; i++) if ((P[i] < i && Answer_2 < P[i] - i + n + 1) || (P[i] >= i && Answer_2 != P[i] - i)) {Answer_2 = INT_MAX;break;}
}

温馨提示：以上代码可通过魔改跑出 $31$ms的最优解。