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<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);++i)
#define reps(i,n) for(int i=1;i<=(n);++i)
#define repr(i,n) for(int i=(n);i>=0;--i)
#define cinv(n,vec) V vec(n);for(auto&v:vec)cin>>v;
#define int long long
#define itn int
#define pb push_back
#define m_p make_pair
#define m_t make_tuple
#define eb emplace_back
#define p_p(a,b) pb(m_p(a,b))
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define RSORT(a) sort(all(a));reverse(all(a))
#define UNIQUE(a) a.erase(unique(all(a)),a.end())
#define SZ(a) (ll)(a.size())
#define ENDL cout<<endl
#define debg(a) cout<<#a<<" "<<a<<endl;
#define debgp(a) cout<<#a<<" "<<a.fi<<" : "<<a.se<<endl
#define call(a) for(auto i:a)cout<<i<<" ";cout<<endl
#define show(a) for(cont &y:a){for(cont &x:y){cout<<x<<" ";}cout<<endl;}
#define out(a) cout<<(a)<<endl
#define cont const auto
#define fi first
#define se second
#define V vector
#define P pair
#define T tuple
#define VV(type,name,y,x,a) vector<vector<type>> name = vector<vector<type>>(y,vector<type>(x,a))
#define PQ priority_queue
typedef long long ll;
typedef double D;
typedef long double LD;
typedef string str;
typedef vector<ll> VL;
typedef pair<ll,ll> PL;
typedef vector<pair<ll,ll>> VP;
typedef tuple<ll,ll,ll> T3;
typedef tuple<ll,ll,ll,ll> T4;
const long long INF = 1LL<<60;
const long long MOD = 1e9+7;
void Yes(){cout<<"Yes"<<endl;}
void No(){cout<<"No"<<endl;}
void YN(bool b){cout<<(b?"Yes":"No")<<endl;}
template<class T>inline T gcd(T a,T b){if(b==0)return a; return(gcd(b,a%b));}
template<class T>inline T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T>inline bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int h,w,d,q;
  cin>>h>>w>>d;
  map<ll,PL> a;
  reps(i,h){
    reps(j,w){
      int in;
      cin>>in;
      a[in]=m_p(i,j);
    }
  }
  cin>>q;
  VL l(q),r(q);
  rep(i,q)cin>>l[i]>>r[i];

  rep(i,q){
    if(l[i]==r[i]){out(0);continue;}
    PL now=a[l[i]];
    ll idx=l[i]+d;
    ll ans=0;
    while(1){
      PL next=a[idx];
      ll res=abs(next.fi-now.fi)+abs(next.se-now.se);
      ans+=res;
      if(idx==r[i])break;
      now=next;
      idx+=d;
    }
    out(ans);
  }

  
  
  
  
  return 0;
}