M - Sum

XXX is puzzled with the question below:

1, 2, 3, ..., n (1<=n<=400000) are placed in a line. There are m (1<=m<=1000) operations of two kinds.

Operation 1: among the x-th number to the y-th number (inclusive), get the sum of the numbers which are co-prime with p( 1 <=p <= 400000).
Operation 2: change the x-th number to c( 1 <=c <= 400000).

For each operation, XXX will spend a lot of time to treat it. So he wants to ask you to help him.

Input

There are several test cases.
The first line in the input is an integer indicating the number of test cases.
For each case, the first line begins with two integers --- the above mentioned n and m.
Each the following m lines contains an operation.
Operation 1 is in this format: "1 x y p".
Operation 2 is in this format: "2 x c".

Output

For each operation 1, output a single integer in one line representing the result.

Sample Input

1
3 3
2 2 3
1 1 3 4
1 2 3 6

Sample Output

7
0

用容斥原理求出不是互质的的数的和,在用等差求和减去不是互质的和,在对改动的数字进行特判

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转载自www.cnblogs.com/wzl19981116/p/9347085.html
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