PivotOJ

PROGRAM

시간 제한: 2000ms메모리 제한: 256MB출처: COCI 2009-2010BOJ 2900

문제

Mirko is trying to debug a piece of his code. First he creates an array of N integers and fills it with zeros. Then he repeatedly calls the following procedure (he is such a good coder he coded it in both C++ and Pascal):

void something(int jump) {
    int i = 0;
    while (i < N) {
        a[i] = a[i] + 1;
        i = i + jump;
    }
}

As you can see, this procedure increases by one all elements in the array whose indices are divisible by jump.

Mirko calls the procedure exactly K times, using the sequence X1 X2 X3... Xk as arguments.

After this, Mirko has a list of Q special parts of the array he needs to check to verify that his code is working as it should be. Each of this parts is defined by two numbers, L and R (L ≤ R) the left and right bound of the special part. To check the code, Mirko must compute the sum of all elements of seq between and including L and R. In other words seq[L] + seq[L+1] + seq[L+2] + ... + seq[R]. Since he needs to know the answer in advance in order to check it, he asked you to help him.

입력

The first line of input contains two integers, N (1 ≤ N ≤ 106), size of the array, and K (1 ≤ K ≤ 106), number of calls to something Mirko makes.

The second line contains K integers: X1X2X3 ... Xk, arguments passed to the procedure. (1 ≤ Xi < N).

Next line contains one integer Q (1 ≤ Q ≤ 106), number of special parts of the array Mirko needs to check.

Next Q lines contain two integers each Li and Ri(0 ≤ Li≤ Ri < N), bounds of each special part.

출력

The output should contain exactly Q lines. The ith line should contain the sum of elements seq[Li] + seq[Li+1] + seq[Li+2] + ... + seq[Ri].

예제

예제 1

입력
10 4
1 1 2 1
3
0 9
2 6
7 7
출력
35
18
3

예제 2

입력
11 3
3 7 10
3
0 10
2 6
7 7
출력
8
2
1

예제 3

입력
1000000 6
12 3 21 436 2 19
2
12 16124
692 29021
출력
16422
28874
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