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: Accepted)

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
typedef pair<int, int> P;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int H, W, D;
  cin >> H >> W >> D;
  int A[H][W];
  P num[H*W+1];
  FOR(i,0,H) {
    FOR(j,0,W) {
      cin >> A[i][j];
      num[A[i][j]] = P(i, j);
    }
  }
  int cost[D+1][(H*W+D-1)/D+1];
  FOR(i,1,D+1) {
    int x = i;
    cost[i][0] = 0;
    int cnt = 1;
    while(x + D <= H * W) {
      // x -> x + D
      cost[i][cnt] = cost[i][cnt-1] + abs(num[x].first-num[x+D].first) + abs(num[x].second-num[x+D].second);
      x += D;
      cnt++;
    }
  }
  int Q; cin >> Q;
  
  FOR(i,0,Q) {
    int L, R; cin >> L >> R;
    int x = (L - 1) / D;
    int st = L - x * D;
    //cout << x << st << (R-st)/D<< endl;
    cout << cost[st][(R-st)/D] - cost[st][x] << endl;
  }
}