Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns. The square at the i-th row and the j-th column will be called Square (i,j).

The integers from 1 through H×W are written throughout the grid, and the integer written in Square (i,j) is A_{i,j}.

You, a magical girl, can teleport a piece placed on Square (i,j) to Square (x,y) by consuming |x-i|+|y-j| magic points.

You now have to take Q practical tests of your ability as a magical girl.

The i-th test will be conducted as follows:

Initially, a piece is placed on the square where the integer L_i is written.

Let x be the integer written in the square occupied by the piece. Repeatedly move the piece to the square where the integer x+D is written, as long as x is not R_i. The test ends when x=R_i.

Here, it is guaranteed that R_i-L_i is a multiple of D.

For each test, find the sum of magic points consumed during that test.

Constraints

1 \leq H,W \leq 300

1 \leq D \leq H×W

1 \leq A_{i,j} \leq H×W

A_{i,j} \neq A_{x,y} ((i,j) \neq (x,y))

1 \leq Q \leq 10^5

1 \leq L_i \leq R_i \leq H×W

(R_i-L_i) is a multiple of D.

Input

Input is given from Standard Input in the following format:

H W D
A_{1,1} A_{1,2} ... A_{1,W}
:
A_{H,1} A_{H,2} ... A_{H,W}
Q
L_1 R_1
:
L_Q R_Q

Output

For each test, print the sum of magic points consumed during that test.

Output should be in the order the tests are conducted.

Sample Input 1

3 3 2
1 4 3
2 5 7
8 9 6
1
4 8

Sample Output 1

5

4 is written in Square (1,2).

6 is written in Square (3,3).

8 is written in Square (3,1).

Thus, the sum of magic points consumed during the first test is (|3-1|+|3-2|)+(|3-3|+|1-3|)=5.

Sample Input 2

4 2 3
3 7
1 4
5 2
6 8
2
2 2
2 2

Sample Output 2

0
0

Note that there may be a test where the piece is not moved at all, and there may be multiple identical tests.

Sample Input 3

5 5 4
13 25 7 15 17
16 22 20 2 9
14 11 12 1 19
10 6 23 8 18
3 21 5 24 4
3
13 13
2 10
13 13

Sample Output 3

0
5
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <unordered_map>

using std::cin;
using std::cout;
using std::endl;

#define ForIndex(N) for (size_t index = 0; index < (N); index++)
#define ForIndex2D(X, Y) \
    for (size_t i = 0; i < (X); i++) {\
        for (size_t j = 0; j < (Y); j++)

template<typename T>
void input2DArray(T**& outArray, size_t xsize, size_t ysize)
{
    outArray = new T*[xsize];

    for (size_t i = 0; i < xsize; i++)
    {
        outArray[i] = new T[ysize];
        for (size_t j = 0; j < ysize; j++)
        {
            cin >> outArray[i][j];
        }
    }
}

// char は一気に入力できるので特殊化
template<>
void input2DArray(char**& outArray, size_t xsize, size_t ysize)
{
    outArray = new char*[xsize];

    for (size_t index = 0; index < xsize; index++)
    {
        outArray[index] = new char[ysize];
        cin >> outArray[index];
    }
}

template<typename T>
void delete2DArray(T**& array, size_t size)
{
    for (size_t index = 0; index < size; index++)
    {
        delete [] array[index];
    }

    delete [] array;
}

int calcCost(const std::pair<int, int>& c, const std::pair<int, int>& o)
{
    return std::abs(c.first - o.first) + std::abs(c.second - o.second);
}

int main()
{
    int H, W, D;
    cin >> H >> W >> D;

    int** s;
    input2DArray(s, H, W);

    uint32_t Q;
    cin >> Q;

    int** lr;
    input2DArray(lr, Q, 2);

    // index map を作っておく
    std::unordered_map<int, std::pair<int, int>> mapMemo;
    ForIndex2D(H, W)
    {
        mapMemo[ s[i][j] ] = std::make_pair(i, j);
    }}

    // 完全なコストマップを作る
    std::unordered_map<int, int> costMemo;
    const auto searchIndex = [](int l, int r) {
        return 1000 * r + l;
    };

    const int maxSize = H * W;
    for (int l = 1; l <= maxSize; l++)
    {
        int prevIndex = searchIndex(l, l);
        costMemo[prevIndex] = 0; // l が同じ時はコストゼロ

        for (int r = l + D; r <= maxSize; r += D)
        {
            const auto& currentPair = mapMemo[r];
            const auto& prevPair = mapMemo[r - D];

            const int newIndex = searchIndex(l, r);

            // 1つ手前の
            costMemo[newIndex] =
                costMemo[prevIndex] + calcCost(currentPair, prevPair);

            prevIndex = newIndex;
        }
    }

    ForIndex(Q)
    {
        cout << costMemo[searchIndex(lr[index][0], lr[index][1])] << endl;
    }

    // delete2DArray(lr, Q);
    // delete2DArray(s, H);
    return 0;
}