#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
typedef vector<ll> vi;
typedef vector<vi> vvi;
const string Yes = "Yes";
const string No = "No";
const string YES = "YES";
const string NO = "NO";
const string abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ll MOD = 1000000007;
const ll mod = 998244353;
const long long INF = 1LL << 60;
const long double PI = 3.141592653589793;
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define YESNO(T)               \
    if (T)                     \
    {                          \
        cout << "YES" << endl; \
    }                          \
    else                       \
    {                          \
        cout << "NO" << endl;  \
    }
#define yesno(T)               \
    if (T)                     \
    {                          \
        cout << "yes" << endl; \
    }                          \
    else                       \
    {                          \
        cout << "no" << endl;  \
    }
#define YesNo(T)               \
    if (T)                     \
    {                          \
        cout << "Yes" << endl; \
    }                          \
    else                       \
    {                          \
        cout << "No" << endl;  \
    }
#define inV(vec)                        \
    for (ll i = 0; i < vec.size(); i++) \
        cin >> vec[i];
#define outV(vec)                       \
    for (ll i = 0; i < vec.size(); i++) \
    {                                   \
        cout << vec[i] << " ";          \
    }                                   \
    cout << endl;
#define print(s) cout << s << endl;
template <typename T>
inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template <typename T>
inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }

// 素因数分解
vector<pair<long long, long long>> prime_factorize(long long N)
{
    vector<pair<long long, long long>> res;
    for (long long a = 2; a * a <= N; ++a)
    {
        if (N % a != 0)
            continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0)
        {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1)
        res.push_back({N, 1});
    return res;
}

using mint = modint998244353;

ll mypow(ll a, ll b)
{
    ll ans = 1;
    for (int i = 0; i < b; i++)
    {
        ans *= a;
    }
    return ans;
}

struct Comb
{
    vector<vector<long long>> com; // 前計算の結果を保存
    long long p;                   // p は素数である必要がある
    vector<ll> fact;               // 階乗
    Comb(long long _p) : p(_p)
    {
        init(p);
    }

    void init(long long p)
    { // 動的計画法で前処理
        com.assign(p, vector<long long>(p));
        com[0][0] = 1;
        for (int i = 1; i < p; i++)
        {
            com[i][0] = 1;
            for (int j = i; j > 0; j--)
            {
                com[i][j] = (com[i - 1][j - 1] + com[i - 1][j]) % p;
            }
        }
    }

    long long nCk(long long n, long long k)
    {
        long long ret = 1;
        while (n > 0)
        { // 下から一桁ずつ計算する
            int ni = n % p;
            int ki = k % p;
            ret *= com[ni][ki];
            ret %= p;
            n /= p;
            k /= p;
        }
        return ret;
    }

    ll modpow(ll a, ll n)
    {
        if (a == 0)
        {
            return 0;
        }
        ll res = 1;
        while (n > 0)
        {
            if (n & 1)
                res = res * a % p;
            a = a * a % p;
            n >>= 1;
        }
        return res;
    }

    void fac(ll N)
    {
        fact.pb(1);
        for (int i = 1; i <= N; i++)
        {
            ll ppp = i;
            while (ppp % p == 0)
            {
                ppp /= p;
            }
            fact.pb(fact[i - 1] * (ppp));
            fact[i] %= p;
        }
    }

    ll Div(ll a, ll b)
    {
        return (a * modpow(b, p - 2)) % p;
    }

    ll nCk2(ll n, ll r)
    {
        if (n < r)
        {
            return 0;
        }

        if (r < 0)
        {
            return 0;
        }

        if (n < 0)
        {
            return 0;
        }
        return Div(fact[n], fact[r] * fact[n - r] % p);
    }
};

// ダイクストラ
struct Edge
{
    long long to;
    long long cost;
};
void dijkstra(const vector<vector<Edge>> &G, int s, vector<long long> &dis)
{
    int N = G.size();
    dis.resize(N, INF);
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq; // 「仮の最短距離, 頂点」が小さい順に並ぶ
    dis[s] = 0;
    pq.emplace(dis[s], s);
    while (!pq.empty())
    {
        pair<long long, int> p = pq.top();
        pq.pop();
        int v = p.second;
        if (dis[v] < p.first)
        { // 最短距離で無ければ無視
            continue;
        }
        for (auto &e : G[v])
        {
            if (dis[e.to] > dis[v] + e.cost)
            { // 最短距離候補なら priority_queue に追加
                dis[e.to] = dis[v] + e.cost;
                pq.emplace(dis[e.to], e.to);
            }
        }
    }
}

int main()
{
    ll H, W, N;
    cin >> H >> W >> N;
    if(N==0){
        print(H+W-2);
        exit(0);
    }
    vector<ll> A(2*N), B(2*N);
    for (int i = 0; i < N; i++)
    {
        cin >> A[2*i] >> B[2*i] >> A[2*i+1] >> B[2*i+1];
    }
    vector<vector<Edge>>Gra(2*N+2);
    for(int i=0;i<2*N;i++){
        for(int j=0;j<2*N;j++){
            if(i%2==0 && i+1==j){
                Gra[i].pb({j,1});
            }else if(i!=j){
                Gra[i].pb({j,abs(A[i]-A[j])+abs(B[i]-B[j])});
            }
        }
        Gra[i].pb({2*N,A[i]+B[i]-2});
        Gra[i].pb({2*N+1,H+W-A[i]-B[i]});
        Gra[2*N].pb({i,A[i]+B[i]-2});
        Gra[2*N+1].pb({i,H+W-A[i]-B[i]});
    }
    vector<ll>ans(2*N+2,INF);
    ans[2*N]=0;
    dijkstra(Gra,2*N,ans);
    print(ans[2*N+1]);
}