#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 200001;
template<class T, int SZ> struct Seg {
T seg[2*SZ], MN = -MOD;
Seg() {
F0R(i,2*SZ) seg[i] = -MOD;
}
T comb(T a, T b) { return max(a,b); } // easily change this to min or max
void upd(int p, T value) { // set value at position p
for (seg[p += SZ] = value; p > 1; p >>= 1)
seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
}
void build() {
F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
}
T query(int l, int r) { // sum on interval [l, r]
T res1 = MN, res2 = MN; r++;
for (l += SZ, r += SZ; l < r; l >>= 1, r >>= 1) {
if (l&1) res1 = comb(res1,seg[l++]);
if (r&1) res2 = comb(seg[--r],res2);
}
return comb(res1,res2);
}
};
Seg<int,1<<20> S;
int N,K;
vpi p;
map<int,int> m;
vi rm, t;
void input() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> K; t.resize(K); p.resize(N);
F0R(i,N) {
cin >> p[i].f >> p[i].s;
m[p[i].f] = m[p[i].s] = 0;
}
F0R(i,K) {
cin >> t[i];
m[t[i]] = 0;
}
for (auto& a: m) {
a.s = sz(rm);
rm.pb(a.f);
}
F0R(i,N) {
p[i].f = m[p[i].f];
p[i].s = m[p[i].s];
}
F0R(i,K) {
t[i] = m[t[i]];
S.upd(t[i],i);
}
}
int main() {
input();
Tree<int> tmp;
vpi T; F0R(i,K) T.pb({t[i],i}); sort(all(T));
sort(all(p),[](pi a, pi b) { return max(a.f,a.s) > max(b.f,b.s); });
ll ans = 0;
for (auto a: p) {
while (sz(T) && T.back().f >= max(a.f,a.s)) tmp.insert(T.back().s), T.pop_back();
int x = S.query(min(a.f,a.s),max(a.f,a.s)-1);
int flip = sz(tmp)-tmp.order_of_key(x);
a.f = rm[a.f], a.s = rm[a.s];
if (x == -MOD) {
if (flip&1) ans += a.s;
else ans += a.f;
} else {
if (flip&1) ans += min(a.f,a.s);
else ans += max(a.f,a.s);
}
// cout << a.f << " " << a.s << " " << x << " " << flip << " " << ans << "\n";
}
cout << ans;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
