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: Wrong Answer)

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

//定数
const long long MOD1=1000000007;
const long long MOD2=998244353;
const long long INF=1152921504606846976;
const long double PI=3.1415926535897932;

//型名省略
#define ll long long
#define ull unsigned long long
#define ld long double
#define pll pair<long long,long long>
#define vl vector<long long>
#define vvl vector<vector<long long>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define umap unordered_map
#define uset unordered_set
#define Lqueue priority_queue<long long>
#define Squeue priority_queue<long long,vector<long long>,greater<long long>>
#define Fi first
#define Se second

//マクロ
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define rbf(a,x) for(auto& a:x)
#define rep(i,n) for(long long i=0;i<(long long)(n);i++)
#define rep2(i,s,n) for(long long i=(s);i<(long long)(n);i++)
#define Maxe(x) *max_element((x).begin(),(x).end())
#define Mine(x) *min_element((x).begin(),(x).end())
#define Size(x) ((long long)(x).size())

//最大公約数
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}

//最小公倍数
long long lcm(long long a,long long b){return a/gcd(a,b)*b;}

//累乗
template<typename Num>
constexpr Num mypow(Num a,ll b){
    if(b==0)return 1;
    if(a==0)return 0;
    Num x=1;
    while(b>0){
        if(b&1)x*=a;
        a*=a;
        b>>=1;
        }
    return x;
}

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

//MODint
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }

    friend constexpr istream& operator >> (istream &is, Fp<MOD>& x){
        long long t;
        is >> t;
        x=t;
        return (is);
    }

    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
using mint = Fp<MOD1>;

    //start
    signed main(){
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        ll h,w,d;
        cin>>h>>w>>d;
        vvl x(d,vl((h*w+d-1)/d,0));
        vvl y(d,vl((h*w+d-1)/d,0));
        rep(i,h)rep(j,w){
            ll a;
            cin>>a;
            x[a%d][a/d]=j;
            y[a%d][a/d]=i;
        }
        vvl cost(d,vl((h*w+d-1)/d,0));
        rep(i,d){
            rep(j,(h*w+d-1)/d-1){
                cost[i][j+1]=abs(x[i][j]-x[i][j+1])+abs(y[i][j]-y[i][j+1])+cost[i][j];
            }
        }
        ll q;
        cin>>q;
        rep(i,q){
            ll l,r;
            cin>>l>>r;
            cout<<cost[r%d][r/d]-cost[l%d][l/d]<<endl;
        }
    }