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E. More Queries to Array...

原题:http://codeforces.com/problemset/problem/266/E

You've got an array, consisting of n integers: a1, a2, ..., an. Your task is to quickly run the queries of two types:

  1. Assign value x to all elements from l to r inclusive. After such query the values of the elements of array al, al + 1, ..., ar become equal to x.

  2. Calculate and print sum , where k doesn't exceed 5. As the value of the sum can be rather large, you should print it modulo 1000000007 (109 + 7).

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105), showing, how many numbers are in the array and the number of queries, correspondingly. The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 109) — the initial values of the array elements.

Then m queries follow, one per line:

  1. The assign query has the following format: "", (1 ≤ l ≤ r ≤ n; 0 ≤ x ≤ 109).

  2. The query to calculate the sum has the following format: "", (1 ≤ l ≤ r ≤ n; 0 ≤ k ≤ 5).

All numbers in the input are integers.

Output

For each query to calculate the sum print an integer — the required sum modulo 1000000007 (109 + 7).

Examples

input

4 5

5 10 2 1

? 1 2 1

= 2 2 0

? 2 4 3

= 1 4 1

? 1 4 5

output

25

43

1300

input

3 1

1000000000 1000000000 1000000000

? 1 3 0

output

999999986


题目大意:

给出一个长度为n的序列。给出m个操作:

  1. 区间[li..ri]赋值为w

  2. 给出(li,ri,k)。求区间[li..ri]的sigma(A[i]*(i-l+1)^k),其中k<=5


个人解法:

想了半天结果是一个套路题……

由于k<=5。用二项式拆分之后,发现最高次幂为5。

也就是说,我们只需要维护A[i],A[i]*i,A[i]*i^2...A[i]*i^5。就可以通过二项式系数解决询问了。

由于需要区间赋值。恰好,我们使用线段树即可。


代码如下:

http://codeforces.com/contest/266/submission/25981137

 
 
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