CF1117G.Recursive Queries

You are given a permutation $p_1, p_2, \dots, p_n$. You should answer $q$ queries. Each query is a pair $(l_i, r_i)$, and you should calculate $f(l_i, r_i)$.

Let's denote $m_{l, r}$ as the position of the maximum in subsegment $p_l, p_{l+1}, \dots, p_r$.

Then $f(l, r) = (r - l + 1) + f(l, m_{l,r} - 1) + f(m_{l,r} + 1, r)$ if $l \le r$ or $0$ otherwise.

给你一个排列 $p_1, p_2, \dots, p_n$。你需要回答 $q$ 个查询。每个查询是一个数对 $(l_i, r_i)$，你需要计算 $f(l_i, r_i)$。

记 $m_{l, r}$ 为子段 $p_l, p_{l+1}, \dots, p_r$ 中最大值的位置。

那么，当 $l \le r$ 时，$f(l, r) = (r - l + 1) + f(l, m_{l,r} - 1) + f(m_{l,r} + 1, r)$；否则 $f(l, r) = 0$。

The first line contains two integers $n$ and $q$ ($1 \le n \le 10^6$, $1 \le q \le 10^6$) — the size of the permutation $p$ and the number of queries.

The second line contains $n$ pairwise distinct integers $p_1, p_2, \dots, p_n$ ($1 \le p_i \le n$, $p_i \neq p_j$ for $i \neq j$) — permutation $p$.

The third line contains $q$ integers $l_1, l_2, \dots, l_q$ — the first parts of the queries.

The fourth line contains $q$ integers $r_1, r_2, \dots, r_q$ — the second parts of the queries.

It's guaranteed that $1 \le l_i \le r_i \le n$ for all queries.

第一行包含两个整数 $n$ 和 $q$（$1 \le n \le 10^6$，$1 \le q \le 10^6$）——分别为排列 $p$ 的长度和查询次数。

第二行包含 $n$ 个两两不同的整数 $p_1, p_2, \dots, p_n$（$1 \le p_i \le n$，当 $i \neq j$ 时 $p_i \neq p_j$）——即排列 $p$。

第三行包含 $q$ 个整数 $l_1, l_2, \dots, l_q$ —— 各查询的左端点。

第四行包含 $q$ 个整数 $r_1, r_2, \dots, r_q$ —— 各查询的右端点。

保证对所有查询均有 $1 \le l_i \le r_i \le n$。

Print $q$ integers — the values $f(l_i, r_i)$ for the corresponding queries.

输出 $q$ 个整数——即各查询对应的 $f(l_i, r_i)$ 的值。

Description of the queries:

1. $f(2, 2) = (2 - 2 + 1) + f(2, 1) + f(3, 2) = 1 + 0 + 0 = 1$;
2. $f(1, 3) = (3 - 1 + 1) + f(1, 2) + f(4, 3) = 3 + (2 - 1 + 1) + f(1, 0) + f(2, 2) = 3 + 2 + (2 - 2 + 1) = 6$;
3. $f(1, 4) = (4 - 1 + 1) + f(1, 2) + f(4, 4) = 4 + 3 + 1 = 8$;
4. $f(2, 4) = (4 - 2 + 1) + f(2, 2) + f(4, 4) = 3 + 1 + 1 = 5$;
5. $f(1, 1) = (1 - 1 + 1) + 0 + 0 = 1$.

查询的描述：

1. $f(2, 2) = (2 - 2 + 1) + f(2, 1) + f(3, 2) = 1 + 0 + 0 = 1$；
2. $f(1, 3) = (3 - 1 + 1) + f(1, 2) + f(4, 3) = 3 + (2 - 1 + 1) + f(1, 0) + f(2, 2) = 3 + 2 + (2 - 2 + 1) = 6$；
3. $f(1, 4) = (4 - 1 + 1) + f(1, 2) + f(4, 4) = 4 + 3 + 1 = 8$；
4. $f(2, 4) = (4 - 2 + 1) + f(2, 2) + f(4, 4) = 3 + 1 + 1 = 5$；
5. $f(1, 1) = (1 - 1 + 1) + 0 + 0 = 1$。