#include <bits/stdc++.h>#define fi first#define se second#define rep( i ,x ,y ) for( int i= x; i<= y ;i++ )#define reb( i ,y ,x ) for( int i= y; i>= x ;i-- )#define mem( a ,x ) memset( a ,x ,sizeof(a))using namespace std;typedef long long ll;typedef pair<int ,int> pii;typedef pair<ll ,ll> pll;typedef pair<string ,int> psi;const int N = 200005;const int inf = 0x3f3f3f3f;ll mp[5000] ,tmp[5000];string s ,t;// check（）匹配次数，是很经典的二分题了bool check( ll x ){ll tm = mp['?'];rep( i ,'a' ,'z' ){if( x*tmp[i] > mp[i])tm -= x*tmp[i] - mp[i];//cout<<"x "<<x<< " i "<<(char)i<< "tm "<<tm<<endl;if( tm <0 )return false;}return true;}int main( ){cin >>s>>t;mem( mp ,0 ); mem( tmp ,0 );int ls = s.size() ,lt = t.size( );rep( i ,0 ,ls-1 )mp[s[i]]++;rep( i ,0 ,lt-1 )tmp[t[i]]++;if( mp['?']==0 )return (cout<<s), 0;ll l = 0 ,r = 1000001 ,ans = 0;while( l<=r ){ll mid = ( l+r ) >>1;if( check(mid) ) l = mid+1 ,ans = mid;else r = mid-1;}//cout<<ans<<endl;rep( i ,'a' ,'z'){mp[i] = ans*tmp[i] - mp[i];}rep( i ,0 ,ls-1 ){if( s[i]=='?' ){int flag = 1;rep( j ,'a' ,'z'){if( mp[j] >0) {cout<<(char)j;mp[j]--; flag = 0;break;}}if( flag )cout<<'a';}else cout<<s[i];}return 0;}

#include <bits/stdc++.h>#define fi first
#define se second
#define rep( i ,x ,y ) for( int i= x; i<= y ;i++ )
#define reb( i ,y ,x ) for( int i= y; i>= x ;i-- )
#define mem( a ,x ) memset( a ,x ,sizeof(a))
using namespace std;typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<ll ,ll> pll;
typedef pair<string ,int> psi;const int N = 200005;
const int inf = 0x3f3f3f3f;int n ,m ,k ,q;
ll li[N] ,ri[N] ,way[N] ,dl[2][N] ,dr[2][N] ,dp[N][5];
ll res ,id[N];
int main( ){res = 0;mem( li ,inf );mem( ri ,-1 );mem( dl ,inf );mem( dr ,inf );mem( dp ,inf );scanf("%d%d%d%d" ,&n ,&m ,&k ,&q );ll tmpl ,tmp;li[1] = ri[1] = 1;rep( i ,1 ,k ){scanf("%lld%lld" ,&tmpl ,&tmp );li[tmpl] = min( li[tmpl] ,tmp );ri[tmpl] = max( ri[tmpl] ,tmp );}rep( i ,1 ,q ){scanf("%lld" ,&way[i]);}sort( way+1 ,way+1+q );int pos1 ,pos2;//找靠近左右最近的梯子 rep( i ,1 ,n ){if( ri[i] == -1 )continue;pos1 = upper_bound( way+1 ,way +1+q ,ri[i] )-way;pos2 = pos1 - 1;// error --2if( pos1 != q+1 ) dr[0][i] = way[pos1];if( pos2 != 0 ) dr[1][i] = way[pos2];pos1 = upper_bound( way+1 ,way +1+q ,li[i] )-way;pos2 = pos1 - 1;if( pos1 != q+1 ) dl[0][i] = way[pos1];if( pos2 != 0 ) dl[1][i] = way[pos2];}ll cnt = 0; tmp = -1;rep( i ,1 ,n ){if( ri[i] == -1 )tmp++;else id[ ++cnt ] = i ,res += ++tmp ,tmp = 0;}//cout<< "dp 1 -- z 2 -- ] 3 -- C 4 -- rz "<<endl;dp[0][1] = dp[0][3] = abs( li[1] - ri[1]);// 初始状态又搞错了 ，初始状态必须从li[1] 开始 rep( i ,1 ,cnt-1 ){ ll tmp1 ,tmp2 ,tmp3 ,tmp4 ,j = id[i] ,k = id[i+1];dp[i][1] =  min(dp[i-1][1] ,dp[i-1][3]) +  abs( li[k] - ri[k]);//cout<<"j ,k "<<j<<" "<<k<<endl;//cout<<"dr dl j "<<dr[0][j]<<" "<<dr[1][j]<<" "<<dl[0][j]<<" "<<dl[1][j]<<endl;//cout<<"dr dl k "<<dr[0][k]<<" "<<dr[1][k]<<" "<<dl[0][k]<<" "<<dl[1][k]<<endl;//cout<<"ri li j "<<ri[j]<<" "<<li[j]<<endl;//cout<<"ri li k "<<ri[k]<<" "<<li[k]<<endl;tmp1 =abs( ri[ j ] - dr[0][j] ) + abs( li[ k ] - dr[0][j] );tmp2 =abs( ri[ j ] - dr[1][j] ) + abs( li[ k ] - dr[1][j] );tmp3 =abs( ri[ j ] - dl[0][k] ) + abs( li[ k ] - dl[0][k] );tmp4 =abs( ri[ j ] - dl[1][k] ) + abs( li[ k ] - dl[1][k] );dp[i][1] += min( min(tmp1 ,tmp2) ,min(tmp3,tmp4));dp[i][2] =  min(dp[i-1][1] ,dp[i-1][3]) +  abs( li[k] - ri[k]);tmp1 =abs( ri[ j ] - dr[0][j] ) + abs( ri[ k ] - dr[0][j] );tmp2 =abs( ri[ j ] - dr[1][j] ) + abs( ri[ k ] - dr[1][j] );tmp3 =abs( ri[ j ] - dr[0][k] ) + abs( ri[ k ] - dr[0][k] );tmp4 =abs( ri[ j ] - dr[1][k] ) + abs( ri[ k ] - dr[1][k] );dp[i][2] += min( min(tmp1 ,tmp2) ,min(tmp3,tmp4));dp[i][3] =  min(dp[i-1][2] ,dp[i-1][4]) + abs( li[k] - ri[k]);tmp1 =abs( li[ j ] - dl[0][j] ) + abs( li[ k ] - dl[0][j] );tmp2 =abs( li[ j ] - dl[1][j] ) + abs( li[ k ] - dl[1][j] );tmp3 =abs( li[ j ] - dl[0][k] ) + abs( li[ k ] - dl[0][k] );tmp4 =abs( li[ j ] - dl[1][k] ) + abs( li[ k ] - dl[1][k] );dp[i][3] += min( min(tmp1 ,tmp2) ,min(tmp3,tmp4));dp[i][4] =  min(dp[i-1][2] ,dp[i-1][4]) + abs( li[k] - ri[k]);tmp1 =abs( li[ j ] - dl[0][j] ) + abs( ri[ k ] - dl[0][j] );tmp2 =abs( li[ j ] - dl[1][j] ) + abs( ri[ k ] - dl[1][j] );tmp3 =abs( li[ j ] - dr[0][k] ) + abs( ri[ k ] - dr[0][k] );tmp4 =abs( li[ j ] - dr[1][k] ) + abs( ri[ k ] - dr[1][k] );dp[i][4] += min( min(tmp1 ,tmp2) ,min(tmp3,tmp4));//cout<<"dp "<<i <<" "<< dp[i][1]<<" "<<dp[i][2]<<" "<<dp[i][3] <<" "<<dp[i][4]<<endl;
    }if(cnt == 1)printf("%lld" ,ri[1]-li[1]); else printf( "%lld" ,res+min( min(dp[cnt-1][1] ,dp[cnt-1][2]) ,min(dp[cnt-1][3] ,dp[cnt-1][4])) ) ;// find shortest way return 0;
}