Willem, Chtholly and Seniorious

【题面】 请你写一种奇怪的数据结构，支持： - $1$ $l$ $r$ $x$ ：将$[l,r]$ 区间所有数加上$x$ - $2$ $l$ $r$ $x$ ：将$[l,r]$ 区间所有数改成$x$ - $3$ $l$ $r$ $x$ ：输出将$[l,r]$ 区间从小到大排序后的第$x$ 个数是的多少(即区间第$x$ 小，数字大小相同算多次，保证 $1\leq$ $x$ $\leq$ $r-l+1$ ) - $4$ $l$ $r$ $x$ $y$ ：输出$[l,r]$ 区间每个数字的$x$ 次方的和模$y$ 的值(即($\sum^r_{i=l}a_i^x$ ) $\mod y$ ) 【输入格式】 这道题目的输入格式比较特殊，需要选手通过$seed$ 自己生成输入数据。 输入一行四个整数$n,m,seed,v_{max}$ （$1\leq $ $n,m\leq 10^{5}$ ,$0\leq seed \leq 10^{9}+7$ $,1\leq vmax \leq 10^{9} $ ） 其中$n$ 表示数列长度，$m$ 表示操作次数，后面两个用于生成输入数据。 数据生成的伪代码如下  其中上面的op指题面中提到的四个操作。 【输出格式】 对于每个操作3和4，输出一行仅一个数。

— Willem... — What's the matter? — It seems that there's something wrong with Seniorious... — I'll have a look...  Seniorious is made by linking special talismans in particular order. After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly. Seniorious has $ n $ pieces of talisman. Willem puts them in a line, the $ i $ -th of which is an integer $ a_{i} $ . In order to maintain it, Willem needs to perform $ m $ operations. There are four types of operations: - $ 1\ l\ r\ x $ : For each $ i $ such that $ l<=i<=r $ , assign $ a_{i}+x $ to $ a_{i} $ . - $ 2\ l\ r\ x $ : For each $ i $ such that $ l<=i<=r $ , assign $ x $ to $ a_{i} $ . - $ 3\ l\ r\ x $ : Print the $ x $ -th smallest number in the index range $ [l,r] $ , i.e. the element at the $ x $ -th position if all the elements $ a_{i} $ such that $ l<=i<=r $ are taken and sorted into an array of non-decreasing integers. It's guaranteed that $ 1<=x<=r-l+1 $ . - $ 4\ l\ r\ x\ y $ : Print the sum of the $ x $ -th power of $ a_{i} $ such that $ l<=i<=r $ , modulo $ y $ , i.e. .

The only line contains four integers $ n,m,seed,v_{max} $ ( $ 1<=n,m<=10^{5},0<=seed<10^{9}+7,1<=vmax<=10^{9} $ ). The initial values and operations are generated using following pseudo code: ``` def rnd(): ret = seed seed = (seed * 7 + 13) mod 1000000007 return ret for i = 1 to n: a[i] = (rnd() mod vmax) + 1 for i = 1 to m: op = (rnd() mod 4) + 1 l = (rnd() mod n) + 1 r = (rnd() mod n) + 1 if (l > r): swap(l, r) if (op == 3): x = (rnd() mod (r - l + 1)) + 1 else: x = (rnd() mod vmax) + 1 if (op == 4): y = (rnd() mod vmax) + 1 ``` Here $ op $ is the type of the operation mentioned in the legend.

For each operation of types $ 3 $ or $ 4 $ , output a line containing the answer.

10 10 7 9

2
1
0
3

10 10 9 9

1
1
3
3

In the first example, the initial array is $ {8,9,7,2,3,1,5,6,4,8} $ . The operations are: - $ 2\ 6\ 7\ 9 $ - $ 1\ 3\ 10\ 8 $ - $ 4\ 4\ 6\ 2\ 4 $ - $ 1\ 4\ 5\ 8 $ - $ 2\ 1\ 7\ 1 $ - $ 4\ 7\ 9\ 4\ 4 $ - $ 1\ 2\ 7\ 9 $ - $ 4\ 5\ 8\ 1\ 1 $ - $ 2\ 5\ 7\ 5 $ - $ 4\ 3\ 10\ 8\ 5 $