// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



typedef long long ll;

#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)

#define ALL(v) (v).begin(),(v).end()

template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}

template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}

template<class T>bool chmin(T&a,const T&b){return a>b?(a=b,1):0;}

template<class T>bool chmax(T&a,const T&b){return a<b?(a=b,1):0;}



const double PI = acos(-1);

ll nextLong() { ll x; scanf("%lld", &x); return x;}

const int MAX_N = 200000 + 10;

vector<pair<ll, ll>> p(MAX_N);



void normalize(vector<pair<ll,ll>>& v, int N) {

  vector<pair<ll,int> > t(N);

  {

    REP(i, N) {

      t[i].first = v[i].first;

      t[i].second = i;

    }

    sort(ALL(t));

    int r = 0;

    REP(i, N) {

      r += (i > 0 && t[i - 1].first != t[i].first);

      v[t[i].second].first = r;

    }

  }

  {

    REP(i, N) {

      t[i].first = v[i].second;

      t[i].second = i;

    }

    sort(ALL(t));

    int r = 0;

    REP(i, N) {

      r += (i > 0 && t[i - 1].first != t[i].first);

      v[t[i].second].second = r;

    }

  }

  return;

}



struct BITMAX {

	int n;

	int vs[MAX_N];

	BITMAX(int n) : n(n) {

    clear();

  }

  void clear() {

    memset(vs, 0, sizeof(vs));

  }

	void update(int idx, int val) {

		for (int x = idx + 1; x <= n; x += x & -x) vs[x] = max(vs[x], val);

	}

	int get(int idx) {

		int ma = 0;

		for (int x = idx + 1; x > 0; x -= x & -x) ma = max(ma, vs[x]);

		return ma;

	}

	

};

BITMAX bit(MAX_N + 10);





int main2() {

  ll x1 = nextLong();

  ll y1 = nextLong();

  ll x2 = nextLong();

  ll y2 = nextLong();

  int N = (int) nextLong();

  REP(i, N) {

    ll X = nextLong();

    ll Y = nextLong();

    p[i].first = X;

    p[i].second = Y;

  }



  // (x1,y1) - (x2,y2) の矩形の外にある点は除く

  {

    int nn = 0;

    REP(i, N) {

      if (min(x1,x2) <= p[i].first  && p[i].first  <= max(x1,x2) && 

          min(y1,y2) <= p[i].second && p[i].second <= max(y1,y2)) {

        p[nn++] = p[i];

      }

    }

    N = nn;

  }





  if (x1 == x2 && y1 == y2) {

    printf("0\n");

    return 0;

  }

  if (x1 == x2) {

    if (y1 > y2) swap(y1, y2);

    double ans = 100 * (double)(y2 - y1);

    REP(i, N) {

      if (p[i].first == x1 && y1 < p[i].second && p[i].second < y2)

        ans += (-20 + 10 * PI);

    }

    printf("%.12f\n", ans);

    return 0;

  }

  if (y1 == y2) {

    if (x1 > x2) swap(x1, x2);

    double ans = 100 * (double)(x2 - x1);

    REP(i, N) {

      if (p[i].second == y1 && x1 < p[i].first && p[i].first < x2)

        ans += (-20 + 10 * PI);

    }

    printf("%.12f\n", ans);

    return 0;

  }



  if (x1 > x2) { swap(x1, x2); REP(i, N) p[i].first *= -1; }

  if (y1 > y2) { swap(y1, y2); REP(i, N) p[i].second *= -1; }

  normalize(p, N);

  

  sort(p.begin(), p.begin() + N);

  bit.clear();

  // pv(p.begin(), p.begin() + N);

  int ma = 0;

  REP(i, N) {

    int b = 1 + bit.get(p[i].second);

    bit.update(p[i].second, b);

    chmax(ma, b);

  }

  double ans = (x2 - x1) * 100 + (y2 - y1) * 100 + ma * (-20 + 5 * PI);

  if (min(x2-x1, y2-y1) + 1 == ma) {

    ans = (x2 - x1) * 100 + (y2 - y1) * 100 + (ma-1) * (-20 + 5 * PI) + (-20 + 10 * PI);

  }

  printf("%.12f\n", ans);

  

  return 0;

}



int main() {

  for (;!cin.eof();cin>>ws) main2();

  return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.2140211890953692x):
#include <bits/stdc++.h>

using namespace std;



typedef long long ll;

#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)

#define ALL(v) (v).begin(),(v).end()

template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}

template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}

template<class T>bool chmin(T&a,const T&b){return a>b?(a=b,1):0;}

template<class T>bool chmax(T&a,const T&b){return a<b?(a=b,1):0;}



const double PI = acos(-1);

ll nextLong() { ll x; scanf("%lld", &x); return x;}

const int MAX_N = 200000 + 10;

vector<pair<ll, ll>> p(MAX_N);



const ll INF = 1LL << 58;

ll dp[MAX_N];

int lis(const vector<pair<ll,ll>>& A, int N) {

  fill(dp, dp + N, INF);

  REP(i, N) {

    int pos = lower_bound(dp, dp + N, A[i].second) - dp;

    dp[pos] = A[i].second;

  }

  return lower_bound(dp, dp + N, INF) - dp;

}



int main2() {

  ll x1 = nextLong();

  ll y1 = nextLong();

  ll x2 = nextLong();

  ll y2 = nextLong();

  int N = (int) nextLong();

  REP(i, N) {

    ll X = nextLong();

    ll Y = nextLong();

    p[i].first = X;

    p[i].second = Y;

  }



  // (x1,y1) - (x2,y2) の矩形の外にある点は除く

  {

    int nn = 0;

    REP(i, N) {

      if (min(x1,x2) <= p[i].first  && p[i].first  <= max(x1,x2) && 

          min(y1,y2) <= p[i].second && p[i].second <= max(y1,y2)) {

        p[nn++] = p[i];

      }

    }

    N = nn;

  }





  if (x1 == x2 && y1 == y2) {

    printf("0\n");

    return 0;

  }

  if (x1 == x2) {

    if (y1 > y2) swap(y1, y2);

    double ans = 100 * (double)(y2 - y1);

    REP(i, N) {

      if (p[i].first == x1 && y1 < p[i].second && p[i].second < y2)

        ans += (-20 + 10 * PI);

    }

    printf("%.12f\n", ans);

    return 0;

  }

  if (y1 == y2) {

    if (x1 > x2) swap(x1, x2);

    double ans = 100 * (double)(x2 - x1);

    REP(i, N) {

      if (p[i].second == y1 && x1 < p[i].first && p[i].first < x2)

        ans += (-20 + 10 * PI);

    }

    printf("%.12f\n", ans);

    return 0;

  }



  if (x1 > x2) { swap(x1, x2); REP(i, N) p[i].first *= -1; }

  if (y1 > y2) { swap(y1, y2); REP(i, N) p[i].second *= -1; }

  

  sort(p.begin(), p.begin() + N);

  {

    ll mi = p[0].second;

    REP(i, N) chmin(mi, p[i].second);

    REP(i, N) p[i].second -= mi;

  }

  int ma = lis(p, N);

  double ans = (x2 - x1) * 100 + (y2 - y1) * 100 + ma * (-20 + 5 * PI);

  if (min(x2-x1, y2-y1) + 1 == ma) {

    ans = (x2 - x1) * 100 + (y2 - y1) * 100 + (ma-1) * (-20 + 5 * PI) + (-20 + 10 * PI);

  }

  printf("%.12f\n", ans);

  

  return 0;

}



int main() {

  for (;!cin.eof();cin>>ws) main2();

  return 0;

}