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: Wrong Answer)

//#pragma once
#include <stdio.h>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <sstream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <numeric>
#include<bitset>

using namespace std;
#define modd 1000000007


typedef pair<long long, long long> pl;



#define ll long long




/*
class UnionFind {
public:
	vector <ll> par;
	vector <ll> siz;
	vector <ll> gr;

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) ,gr(sz_){
		for (ll i = 0; i < sz_; ++i) par[i] = i;
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;

		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}



	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}


};*/


long long N, M, K, a, b, c, d, e, H, W, L;
long long A[200005] = {};
long long B[500][500] = {};
long long C[205][205] = {};
long long D[200005] = {};
long long E[200005] = {};
bool f, ff;
string S;
set <long long>sll;
pl bufpl;
vector <long long>vl;
vector <pl>vpl;
vector <string> vs;
set<ll> llset;

/*struct ST
{
	long long Kati;
	int bangou;

	bool operator<(const ST& another) const
	{
		return Kati < another.Kati;//比較
	};

};



vector <ST> vst;
ST st[200005];
ST bufst;
bitset<5000> bits;*/


long long modinv(long long aa, long long mm) {
	long long bb = mm, uu = 1, vv = 0;
	while (bb) {
		long long tt = aa / bb;
		aa -= tt * bb; swap(aa, bb);
		uu -= tt * vv; swap(uu, vv);
	}
	uu %= mm;
	if (uu < 0) uu += mm;
	return uu;
}


bool fs[2005][2005] = {};
char cc[1000000] = {};
pl pld[100000];

int main() {


	cin >> H>>W>>d;

	for (int i = 1; i <= H;i++) {
		for (int j = 1; j <= W; j++) {
			cin >> a;
			pld[a].first = i;
			pld[a].second = j;
		}
	}
	c = 0;
	cin >> e;
	for (int i = 0; i < e; i++) {
		cin >> a >> b;
		while (a < b) {

			c += abs(pld[a].first - pld[a + d].first) + abs(pld[a].second - pld[a + d].second);

			a += d;
		}
	}
	


	cout << c << endl;

	return 0;

}