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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<algorithm>     //sort,二分探索,など
#include<bitset>        //固定長bit集合
#include<cmath>         //pow,logなど
#include<complex>       //複素数
#include<deque>         //両端アクセスのキュー
#include<functional>    //sortのgreater
#include<iomanip>       //setprecision(浮動小数点の出力の誤差)
#include<iostream>      //入出力
#include<iterator>      //集合演算(積集合,和集合,差集合など)
#include<map>           //map(辞書)
#include<numeric>       //iota(整数列の生成),gcdとlcm(c++17)
#include<queue>         //キュー
#include<set>           //集合
#include<stack>         //スタック
#include<string>        //文字列
#include<unordered_map> //イテレータあるけど順序保持しないmap
#include<unordered_set> //イテレータあるけど順序保持しないset
#include<utility>       //pair
#include<vector>        //可変長配列

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvll;
typedef vector<vector<string>> vvs;
typedef vector<vector<bool>> vvb;
const ll MOD = 1000000007;
const ll INF = 1000000000000000000;
#define rep(i,n) for(int i=0;i<n;i++)
#define repl(i,s,e) for(int i=s;i<e;i++)
#define reple(i,s,e) for(int i=s;i<=e;i++)
#define revrep(i,n) for(int i=n-1;i>=0;i--)
#define all(x) (x).begin(),(x).end()

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int H, W, D;
    cin >> H >> W >> D;

    int N = H * W;

    vvi A(H, vi(W));
    rep(i, H)
    {
        rep(j, W)
        {
            cin >> A[i][j];
            A[i][j]--;
        }
    }

    vector<pair<int, int>> pos(N);
    rep(i, H)
    {
        rep(j, W)
        {
            pos[A[i][j]] = make_pair(i, j);
        }
    }

    //コストを累積和で計算
    vvll cost(D, vll(N / D + 1));
    rep(d, D)
    {
        for (int c = 0; c < N; c++)
        {
            int next = (c + 1) * D + d;
            if (next >= N) break;
            auto p = pos[next - D];
            auto q = pos[next];
            cost[d][c + 1] = cost[d][c] + (abs(p.first - q.first) + abs(p.second - q.second));
        }
    }

    int Q;
    cin >> Q;

    rep(i, Q)
    {
        int L, R;
        cin >> L >> R;
        L--;
        R--;

        int d = L % D;
        int l = L / D;
        int r = R / D;

        cout << cost[d][r] - cost[d][l] << endl;
    }

    return 0;
}