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 <string>
#include <unordered_map>
#include <stdio.h>
#include <list>
#include <vector>
#include <algorithm>
#include <stack>
#include <climits>
#include <cmath>
#include <map>
#include <queue>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; ++i)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

int main()
{
	int h,w,d;
	cin >> h >> w >> d;
	vector<vector<int>> A(h,vector<int>(w));
	rep(i,h)
	{
		rep(j,w)
		{
		cin>>A[i][j];
		A[i][j]--;
		}
	}
	//表をゼロオリジンにした
	vector<int> dist(h*w-d);

	for(int i=0;i<h*w-d;i++)
	{
		int x1,y1,x2,y2;
		for(int j=0;j<h;j++)
			for(int k=0;k<w;k++)
			{
				if(i==A[j][k])
				{
					x1=j;y1=k;
				}
				if(i+d == A[j][k])
				{
					x2=j;y2=k;
				}
			}
		dist[i]=abs(x1-x2)+abs(y1-y2);
		//cout<<dist[i]<<endl;

	}
	//ｄ個の一次元累積和をとる
	vector<vector<int>> distsum(d,vector<int> ( 1 ) );
	rep(i,d)
	{
		distsum[i][0]=0;
	}
	rep(i,h*w-d)
	{
		distsum[i%d].push_back(distsum[i%d][i/d]+dist[i]);
	}	
	//cout<<distsum[3].size()<<endl;
	//cout<<distsum[1].size()<<endl;
	//rep(i,4)cout<<distsum[1][i]<<endl;


	

	int q;cin>>q;
	rep(iii,q)
	{
		int a,b;
		cin >> a >> b;
		a--;
		b--;
		//cout<<b%d<<':'<<b/d<<endl;
		//cout<<a%d<<':'<<a/d<<endl;
		cout<<distsum[b%d][b/d]-distsum[a%d][a/d]<<endl;
	}

	
	
	return 0;
}