stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <iostream>

#include <string.h>

#include <stdio.h>

#include <queue>

#include <vector>

#include <map>

#include <algorithm>

using namespace std;

const int Max=3e5+100;

int get_root(int *f, int root){

    return f[root]==root?root:f[root]=get_root(f, f[root]);

}

void merge(int *f, int x,int y){

    int i=get_root(f, x);

    int j=get_root(f, y);

    f[i]=j;

}

int f[Max], g[Max];

void init(int n){

    for(int a=0; a<=n; a++)f[a]=a, g[a]=a;

}

int vis[Max], num[Max];

vector<int> vec[Max];

map<int, int>ma;

int main(){

    int n, m, k, u, v;

    scanf("%d%d%d", &n, &m, &k);

    init(n);

    for(int a=0; a<m; a++){

        scanf("%d%d", &u, &v);

        merge(f, u, v);

    }

    for(int a=0; a<k; a++){

        scanf("%d%d", &u, &v);

        merge(g, u, v);

    }

    for(int a=1; a<=n; a++)vec[get_root(f, a)].push_back(a);

    for(int a=0; a<=n; a++){

        ma.clear();

       // memset(num, 0, sizeof(num));

        for(int b=0; b<vec[a].size(); b++){

            ma[get_root(g, vec[a][b])]++;

            //num[get_root(g, vec[a][b])]++;

        }

        for(int b=0; b<vec[a].size(); b++){

            vis[vec[a][b]]+=ma[get_root(g, vec[a][b])];

        }

    }

    for(int a=1; a<=n; a++)printf("%d%c", vis[a], a==n?'\n':' ');

    return  0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

struct hi{

	ll a;int b;

} cnt[200005];

int id[200005],id2[200005],p[200005]

,n,k,l,i,j,m;



int find(int x){

	while (x!=id[x]){

		id[x]=id[id[x]];

		x=id[x];

	}

	return x;

}



int find2(int x){

	while (x!=id2[x]){

		id2[x]=id2[id2[x]];

		x=id2[x];

	}

	return x;

}



void connect(int x,int y){

	x=find(x);y=find(y);

	if (x!=y) id[x]=y;

}



void connect2(int x,int y){

	x=find2(x);y=find2(y);

	if (x!=y) id2[x]=y;

}



bool cmp(hi a,hi b){

	return a.a<b.a;

}



int main(){

	scanf("%d%d%d",&n,&k,&l);

	for (i=1;i<=n;i++){

		id[i]=i;id2[i]=i;

	}

	for (i=1;i<=k;i++){

		scanf("%d%d",&j,&m);

		connect(j,m);

	}

	for (i=1;i<=l;i++){

		scanf("%d%d",&j,&m);

		connect2(j,m);

	}

	for (i=1;i<=n;i++){

		id[i]=find(id[i]);id2[i]=find2(id2[i]);

		cnt[i].a=ll(id[i])*200001+ll(id2[i]);cnt[i].b=i;

	}

	sort(cnt+1,cnt+n+1,cmp);

	j=0;k=0;

	for (i=1;i<=n;i++){

		if (cnt[i].a!=cnt[i-1].a) j++;

		p[j]++;

		id[cnt[i].b]=j;

	}

	for (i=1;i<n;i++) printf("%d ",p[id[i]]);

	printf("%d\n",p[id[n]]);

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define ull unsigned long long

#define pb push_back

const int maxn=2e5+10;

const int mod=998244353;

int f1[maxn];

int f2[maxn];



int find1(int x)

{

    if(f1[x]==x)

        return x;

    return f1[x]=find1(f1[x]);

}

int find2(int x)

{

    if(f2[x]==x)

        return x;

    return f2[x]=find2(f2[x]);

}



void add1(int x,int y)

{

    int t1=find1(x),t2=find1(y);

    f1[t1]=t2;

}

void add2(int x,int y)

{

    int t1=find2(x),t2=find2(y);

    f2[t1]=t2;

}



int main()

{

    ios::sync_with_stdio(false);

    cin.tie(0);

    int n,k,l;

    cin>>n>>k>>l;

    for(int i=1;i<maxn;i++)

        f1[i]=f2[i]=i;

    while(k--)

    {

        int p,q;

        cin>>p>>q;

        add1(p,q);

    }

    while(l--)

    {

        int r,s;

        cin>>r>>s;

        add2(r,s);

    }

    map<pair<int,int>,int>mp;

    for(int i=1;i<=n;i++)

    {

        mp[{find1(i),find2(i)}]++;

    }

    for(int i=1;i<=n;i++)

    {

        cout<<mp[{f1[i],f2[i]}]<<" ";

    }





}


// Optimized solution to the same problem (faster):
// ====================================

//   author: M_sea

//   website: https://m-sea-blog.com/

// ====================================

#include <bits/stdc++.h>

#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)

#define debug(...) fprintf(stderr,__VA_ARGS__)

#define mp make_pair

using namespace std;

typedef long long ll;



int read() {

	int X=0,w=1; char c=getchar();

	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }

	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();

	return X*w;

}



const int N=200000+10;



int n,k,l;

struct DSU {

	int f[N];

	int find(int x) { return x==f[x]?x:f[x]=find(f[x]); }

	void merge(int x,int y) {

		x=find(x),y=find(y);

		if (x!=y) f[x]=y;

	}

} A,B;

map<pair<int,int>,int> M;



int main() {

	n=read(),k=read(),l=read();

	for (int i=1;i<=n;++i) A.f[i]=B.f[i]=i;

	for (int i=1;i<=k;++i) A.merge(read(),read());

	for (int i=1;i<=l;++i) B.merge(read(),read());

	for (int i=1;i<=n;++i) ++M[mp(A.find(i),B.find(i))];

	for (int i=1;i<=n;++i) printf("%d ",M[mp(A.find(i),B.find(i))]);

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define rank tipa_rank

#define pb push_back

#define mp make_pair

#define ll long long

#define ull unsigned long long

#define INF 1000000000

#define EPS 1e-6

#define free tipa_free

#define EPS2 1e-8

using namespace std;

const ll MOD = INF + 7;

using namespace std;

string s;

string ans;

int n , k;

deque<pair<char , int> > v;

bool swapped = false;

int main()

{

	//freopen("output.txt" , "w" , stdout);

	cin >> n >> k;

	cin >> s;

	for(int i = 0; i < s.size(); i++)

	{

		if(v.size() && v.back().first == s[i])

			v.back().second++;

		else

			v.push_back(mp(s[i] , 1));

	}

	if(n == 1)

	{

		cout << char(((s[0] - 'A') ^ (k % 2)) + 'A');

		return 0;

	}

	/*for(int i = 0; i < v.size(); i++)

		cout << v[i].first << " " << v[i].second << endl;

	cout << endl;*/

	int cnt = 0;

	while(k--)

	{



		if(v.front().first == 'A' + swapped)

		{

			v.front().second--;

			if(!v.front().second)

				v.pop_front();

			if(v.front().first == 'B' - swapped)

			{

				v.front().second++;

			}

			else

				v.push_front(mp('B' - swapped , 1));

		}

		else

		{

			//v.pop_front();

			v.front().second--;

			if(!v.front().second)

				v.pop_front();

			swapped = !swapped;

			if(v.back().first == swapped + 'A')

				v.back().second++;

			else

				v.push_back(mp(swapped + 'A' , 1));

		}

		/*cout << swapped << endl;

		for(int i = 0; i < v.size(); i++)

			cout << v[i].first << " " << v[i].second << endl;

		cout << endl;*/

		cnt++;

		if(cnt == 500000)

		{

			break;

		}

	}

	if(k % 2 && k > 0 && n % 2)

	{

		if(v.front().second == 1)

		{

			v.pop_front();

			v.front().second++;

		}

		else

		{

			v.front().second--;

			v.push_front(mp('A' + swapped , 1));

		}

	}



	//cout << v[0].first - 'A' << endl;

	for(int i = 0; i < v.size(); i++)

	{

		if(!swapped)

		{

			for(int j = 0; j < v[i].second; j++)

				cout << v[i].first;

		}

		else

		{

			for(int j = 0; j < v[i].second; j++)

				cout << char(((v[i].first - 'A') ^ 1) + 'A');

		}

	}

}


// Optimized solution to the same problem (faster):
#include<cstdio>

#define maxn 200005

int n,k,l,r;

bool rev;

char s[maxn];

int main()

{

	scanf("%d%d%s",&n,&k,s);

	for(int i=0;i<n;i++) s[i]-=65;

	if(k>n*2+1) k=k%2?(n*2+1):(n*2+2);

	r=n-1;

	for(int i=1;i<=k;i++) 

	{

		if(!(s[l]^rev)) s[l]^=1;

		else l=(l+1)%n,r=(r+1)%n,rev^=1;

	}

	for(;;l=(l+1)%n)

	{

		printf("%c",65+(s[l]^rev));

		if(l==r) break;

	}

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define N 200005

char s[N];int n,k,l,r,rev;

int main(){scanf("%d%d%s",&n,&k,s);for(int i=0;i<n;i++) s[i]-=65;if(k>n*2+1) k=k%2?(n*2+1):(n*2+2);r=n-1;

for(int i=1;i<=k;i++)if(!(s[l]^rev)) s[l]^=1;else l=(l+1)%n,r=(r+1)%n,rev^=1;

for(;;l=(l+1)%n){printf("%c",65+(s[l]^rev));if(l==r) break;}}


// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int N=4e5+5;

typedef long long ll;

int n,k;

char s[N];

int num[N];

int st;

int main()

{

    scanf("%d%d",&n,&k);

    scanf("%s",s+1);

    for(int i=1;i<=n;i++)

        num[i]=s[i]-'A';

    st=num[1];

    int p=1;

    int cnt=0;

    while(k--){

        if(st==0){

            st^=1;

            num[p]^=1;

        }

        else{

            p++;

            cnt++;

            st=num[p];

            st^=(cnt&1);

        }

        if(p>n)

            break;

    }

    if(p<=n){

        for(int i=p;i<=n;i++)

            putchar('A'+(num[i]^(cnt&1)));

        int t=p-1,c=(cnt-1)&1;

        while(t--){

            putchar('A'+c);

            c^=1;

        }

    }

    else{

        if((cnt-1)&1)

            for(int i=1;i<=n;i++)

                putchar('A'+(i&1));

        else{

            putchar('A'+(k&1));

            for(int i=1;i<n;i++)

                putchar('A'+(i&1));

        }

    }

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define rank tipa_rank

#define pb push_back

#define mp make_pair

#define ll long long

#define ull unsigned long long

#define INF 1000000000

#define EPS 1e-6

#define free tipa_free

#define EPS2 1e-8

using namespace std;

const ll MOD = INF + 7;

using namespace std;

string s;

string ans;

int n , k;

deque<pair<char , int> > v;

bool swapped = false;

int main()

{

	//freopen("output.txt" , "w" , stdout);

	cin >> n >> k;

	cin >> s;

	for(int i = 0; i < s.size(); i++)

	{

		if(v.size() && v.back().first == s[i])

			v.back().second++;

		else

			v.push_back(mp(s[i] , 1));

	}

	if(n == 1)

	{

		cout << char(((s[0] - 'A') ^ (k % 2)) + 'A');

		return 0;

	}

	/*for(int i = 0; i < v.size(); i++)

		cout << v[i].first << " " << v[i].second << endl;

	cout << endl;*/

	int cnt = 0;

	while(k--)

	{



		if(v.front().first == 'A' + swapped)

		{

			v.front().second--;

			if(!v.front().second)

				v.pop_front();

			if(v.front().first == 'B' - swapped)

			{

				v.front().second++;

			}

			else

				v.push_front(mp('B' - swapped , 1));

		}

		else

		{

			//v.pop_front();

			v.front().second--;

			if(!v.front().second)

				v.pop_front();

			swapped = !swapped;

			if(v.back().first == swapped + 'A')

				v.back().second++;

			else

				v.push_back(mp(swapped + 'A' , 1));

		}

		/*cout << swapped << endl;

		for(int i = 0; i < v.size(); i++)

			cout << v[i].first << " " << v[i].second << endl;

		cout << endl;*/

		cnt++;

		if(cnt == 500000)

		{

			break;

		}

	}

	if(k % 2 && k > 0 && n % 2)

	{

		if(v.front().second == 1)

		{

			v.pop_front();

			v.front().second++;

		}

		else

		{

			v.front().second--;

			v.push_front(mp('A' + swapped , 1));

		}

	}



	//cout << v[0].first - 'A' << endl;

	for(int i = 0; i < v.size(); i++)

	{

		if(!swapped)

		{

			for(int j = 0; j < v[i].second; j++)

				cout << v[i].first;

		}

		else

		{

			for(int j = 0; j < v[i].second; j++)

				cout << char(((v[i].first - 'A') ^ 1) + 'A');

		}

	}

}


// Optimized solution to the same problem (faster):
#include <cstdio>

const int MAXN=220000;

char s[MAXN];

int n, k, p, f;

void solve(int k)

{

	p=1, f=0;

	for (int i=1; i<=k; i++)

		if (f^s[p]=='A') s[p]=f?'A':'B';

		else f^=1, p=p==n?1:p+1;

}

int main()

{

//	freopen("D.in", "r", stdin);

//	freopen("D.out", "w", stdout);

	scanf("%d%d%s", &n, &k, s+1);

	if (k<=2*n)

	{

		solve(k);

		for (int i=p, j=1; j<=n; i=i==n?1:i+1, j++)

		if (!f) putchar(s[i]);

		else putchar(s[i]=='A'?'B':'A');

	}

	else

	{

		solve(2*n);

		if (n%2==0)

			for (int i=1, f=0; i<=n; i++, f^=1) putchar(f?'A':'B');

		else if (s[p]=='A'^(k-2*n)%2==0)

			for (int i=1, f=1; i<=n; i++, f^=1) putchar(f?'A':'B');

		else

		{

			putchar('B');

			for (int i=2, f=0; i<=n; i++, f^=1) putchar(f?'A':'B');

		}

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define Fst first

#define Snd second

#define RG register

#define mp make_pair

#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long LL;

typedef long double LD;

typedef unsigned int UI;

typedef unsigned long long ULL;

template<typename T> inline void read(T& x) {

	char c = getchar();

	bool f = false;

	for (x = 0; !isdigit(c); c = getchar()) {

		if (c == '-') {

			f = true;

		}

	}

	for (; isdigit(c); c = getchar()) {

		x = x * 10 + c - '0';

	}

	if (f) {

		x = -x;

	}

}

template<typename T, typename... U> inline void read(T& x, U& ... y) {

	read(x), read(y...);

}

int n,K;

int G[600010];

char s[200010];

//#define rua

int main() {

//	ios::sync_with_stdio(false);

#ifdef rua

	freopen("GG.in","r",stdin);

#endif

  read(n,K); scanf("%s",s+1);

  for(int i=1;i<=n;++i) G[i]=s[i]-'A';

  int l=1,r=n,t=0;

  for(int i=1;i<=n<<1;++i) {

    if(G[l]^t==0) G[l]^=1;

    else t^=1,++l,G[++r]=t;

    if(i==K) {

      for(int i=l;i<=r;++i) G[i]^=t,printf("%c",G[i]+'A');

      return 0;

    }

  }

  if(n%2&&K%2) G[l]^=1;

  for(int i=l;i<=r;++i) G[i]^=t,printf("%c",G[i]+'A');

  return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

const int MAXN=220000;

char s[MAXN];

int n, k, p, f;

void solve(int k)

{

	p=1, f=0;

	for (int i=1; i<=k; i++)

		if (f^s[p]=='A') s[p]=f?'A':'B';

		else f^=1, p=p==n?1:p+1;

}

int main()

{

//	freopen("D.in", "r", stdin);

//	freopen("D.out", "w", stdout);

	scanf("%d%d%s", &n, &k, s+1);

	if (k<=2*n)

	{

		solve(k);

		for (int i=p, j=1; j<=n; i=i==n?1:i+1, j++)

		if (!f) putchar(s[i]);

		else putchar(s[i]=='A'?'B':'A');

	}

	else

	{

		solve(2*n);

		if (n%2==0)

			for (int i=1, f=0; i<=n; i++, f^=1) putchar(f?'A':'B');

		else if (s[p]=='A'^(k-2*n)%2==0)

			for (int i=1, f=1; i<=n; i++, f^=1) putchar(f?'A':'B');

		else

		{

			putchar('B');

			for (int i=2, f=0; i<=n; i++, f^=1) putchar(f?'A':'B');

		}

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



using ll = long long;



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

#define SIZE(a) ((int)(a).size())



const int N = 200005;



const int mod = 1000*1000*1000 + 7;



int n;

pair<int,int> a[N];

int lf[N], rt[N];

int seg[N];

int f[N];



inline void add(int &x, int y) {

	x += y;

	if (x >= mod) x -= mod;

}



int main() {

	ios::sync_with_stdio(false);

	cin.tie(0);



	cin >> n;

	REP(i, n) cin >> a[i].first >> a[i].second;

	REP(i, n+2) seg[i] = -1;

	sort(a, a + n);



	vector<pair<int,int>> b(n);

	REP(i, n) b[i] = {a[i].second, i};

	sort(b.begin(), b.end());



	int mx = -1;

	REP(i, n) {

		mx = max(mx, b[i].second);

		rt[b[i].second] = mx+1;

	}

	int mn = n;

	for (int i = n-1; i >= 0; --i) {

		mn = min(mn, b[i].second);

		lf[b[i].second] = mn+1;

	}



	REP(i, n) {

		seg[rt[i]] = max(seg[rt[i]], lf[i]);

	}



	int sum = 1;

	f[0] = 1;



	mx = -1;

	int j = 0;



	for (int i = 1; i <= n+1; ++i) {

		mx = max(mx, seg[i-1]);

		while (j < mx) {

			add(sum, mod-f[j]);

			++j;

		}

		add(f[i], sum);

		add(sum, sum);

	}



	cout << f[n+1] << '\n';



	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;



const int MAXN=200005;

const int MOD=1e9+7;

const int INF=0x3f3f3f3f;



int n;

int L[MAXN],R[MAXN];

int dp[MAXN];

struct node

{

	int X,V,id;

	node(){};

	node(int _X,int _V):X(_X),V(_V){};

}nd[MAXN+5];



bool cmp1(const node &a,const node &b){return a.X<b.X;}

bool cmp2(const node &a,const node &b){return a.V<b.V;}

int main(){

	scanf("%d",&n);

	for(int i=1;i<=n;i++)

		scanf("%d %d",&nd[i].X,&nd[i].V);

	sort(nd+1,nd+1+n,cmp2);

	for(int i=1;i<=n;i++)

		nd[i].id=i;

	sort(nd+1,nd+1+n,cmp1);

	int mx=-1,mn=INF;

	for(int i=1;i<=n;i++){

		mx=max(mx,nd[i].id);

		R[i]=mx;

	}

	for(int i=n;i>=1;i--){

		mn=min(mn,nd[i].id);

		L[i]=mn;

	}

	int pos=0,sum=1;

	dp[0]=1;

	for(int i=1;i<=n;i++){

		while(R[pos]<L[i]-1)

			sum=((1ll*sum-1ll*dp[pos])%MOD+MOD)%MOD,pos++;

		dp[i]=sum;

		sum=(1ll*dp[i]+1ll*sum)%MOD;

	}

	int ans=0;

	for(int i=1;i<=n;i++)

		if(R[i]>=n)

			ans=(1ll*ans+1ll*dp[i])%MOD;

	printf("%d\n",ans);

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

const int maxn=2e5+5;

bool ok[maxn],tmp[maxn];

string ss;

int n;

bool check(int left,int end)

{

	memset(ok,0,sizeof(ok));

	ok[left]=true;

	for(int i=left+1;i<=end;i++)

		if((ok[i-1]||ok[i-2])&&ss[i]=='.')

			ok[i]=true;

	return ok[end];

}

void fl()

{

	freopen(".in","r",stdin);

	freopen(".out","w",stdout);

}

int main()

{

	ios::sync_with_stdio(0);

//  fl();

	int a,b,c,d;

	cin>>n>>a>>b>>c>>d;

	cin>>ss;

	ss='#'+ss;

	if(c<d)

	{

		if(!check(b,d))

		{

			cout<<"No"<<endl;

			return 0;

		}

		if(!check(a,c))

		{

			cout<<"No"<<endl;

			return 0;

		}

		cout<<"Yes"<<endl;

		return 0;

	}

	else

	{

		if(!check(b,d))

		{

			cout<<"No"<<endl;

			return 0;

		}

		for(int i=1;i<=n;i++)

			tmp[i]=ok[i];

		for(int i=b;i<=d;i++)

			if(tmp[i])

				if(ss[i-1]=='.'&&ss[i+1]=='.'&&check(i,d))

				{

					cout<<"Yes"<<endl;

					return 0;

				}

		cout<<"No"<<endl;

	}

	return 0;

}


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

#define ff first

#define ss second

typedef long long int ll;

#define int long long int

typedef vector<ll> vi;

typedef pair<int, int> ii; typedef vector<ii> vii;

#define INF numeric_limits<ll>::max()

#define all(x) (x).begin(), (x).end()

#define sz(x) (int)(x.size())

ll powmod(ll a,ll l, ll md){ll res=1;while(l){if(l&1)res=res*a%md;l/=2;a=a*a%md;}return res;}

ll binpow(ll a,ll l){ll res=1;while(l){if(l&1)res=res*a;l/=2;a=a*a;}return res;}

ll invmod(ll a, ll md){return powmod(a,md-2,md);}

const ll inf = 0xFFFFFFFFFFFFFFFL;





signed main() {

	ios_base::sync_with_stdio(false);

	cin.tie(NULL); cout.tie(NULL);

	ll n, a, b, c, d;

	cin>>n>>a>>b>>c>>d;

	string s;

	cin>>s;

	a--;b--;c--;d--;

	bool flag = true;

	if ((a<b && b<c && c<d)|| (a<c && c<b && b<d)){

		for (ll i = a+1 ; i<c;i++){

			if (s[i] == '#' && s[i+1]=='#'){

				flag = false;

				break;

			}



		}

		for (ll i = b+1 ; i<d;i++){

			if (s[i] == '#' && s[i+1]=='#'){

				flag = false;

				break;

			}



		}



	}

	else{

		for (ll i = a+1 ; i<c;i++){

			if (s[i] == '#' && s[i+1]=='#'){

				flag = false;

				break;

			}



		}

		if (flag){

			bool flag1 = false;

			for (ll i = b ; i<=d;i++){

				if (s[i-1] == '.' && s[i+1]=='.'&& s[i]== '.'){

					flag1 = true;

					break;

				}

			}

			if (flag1==false)flag = false;



		}





	}

	if (flag)cout<<"Yes"<<"\n";

	else cout<<"No"<<"\n";

    return 0;

}

 

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

const int maxn=2e5+5;

bool ok[maxn],tmp[maxn];

string ss;

int n;

bool check(int left,int end)

{

	memset(ok,0,sizeof(ok));

	ok[left]=true;

	for(int i=left+1;i<=end;i++)

		if((ok[i-1]||ok[i-2])&&ss[i]=='.')

			ok[i]=true;

	return ok[end];

}

void fl()

{

	freopen(".in","r",stdin);

	freopen(".out","w",stdout);

}

int main()

{

	ios::sync_with_stdio(0);

//  fl();

	int a,b,c,d;

	cin>>n>>a>>b>>c>>d;

	cin>>ss;

	ss='#'+ss;

	if(c<d)

	{

		if(!check(b,d))

		{

			cout<<"No"<<endl;

			return 0;

		}

		if(!check(a,c))

		{

			cout<<"No"<<endl;

			return 0;

		}

		cout<<"Yes"<<endl;

		return 0;

	}

	else

	{

		if(!check(b,d))

		{

			cout<<"No"<<endl;

			return 0;

		}

		for(int i=1;i<=n;i++)

			tmp[i]=ok[i];

		for(int i=b;i<=d;i++)

			if(tmp[i])

				if(ss[i-1]=='.'&&ss[i+1]=='.')

				{

					cout<<"Yes"<<endl;

					return 0;

				}

		cout<<"No"<<endl;

	}

	return 0;

}


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

#define ff first

#define ss second

typedef long long int ll;

#define int long long int

typedef vector<ll> vi;

typedef pair<int, int> ii; typedef vector<ii> vii;

#define INF numeric_limits<ll>::max()

#define all(x) (x).begin(), (x).end()

#define sz(x) (int)(x.size())

ll powmod(ll a,ll l, ll md){ll res=1;while(l){if(l&1)res=res*a%md;l/=2;a=a*a%md;}return res;}

ll binpow(ll a,ll l){ll res=1;while(l){if(l&1)res=res*a;l/=2;a=a*a;}return res;}

ll invmod(ll a, ll md){return powmod(a,md-2,md);}

const ll inf = 0xFFFFFFFFFFFFFFFL;





signed main() {

	ios_base::sync_with_stdio(false);

	cin.tie(NULL); cout.tie(NULL);

	ll n, a, b, c, d;

	cin>>n>>a>>b>>c>>d;

	string s;

	cin>>s;

	a--;b--;c--;d--;

	bool flag = true;

	if ((a<b && b<c && c<d)|| (a<c && c<b && b<d)){

		for (ll i = a+1 ; i<c;i++){

			if (s[i] == '#' && s[i+1]=='#'){

				flag = false;

				break;

			}



		}

		for (ll i = b+1 ; i<d;i++){

			if (s[i] == '#' && s[i+1]=='#'){

				flag = false;

				break;

			}



		}



	}

	else{

		for (ll i = a+1 ; i<c;i++){

			if (s[i] == '#' && s[i+1]=='#'){

				flag = false;

				break;

			}



		}

		if (flag){

			bool flag1 = false;

			for (ll i = b ; i<=d;i++){

				if (s[i-1] == '.' && s[i+1]=='.'&& s[i]== '.'){

					flag1 = true;

					break;

				}

			}

			if (flag1==false)flag = false;



		}





	}

	if (flag)cout<<"Yes"<<"\n";

	else cout<<"No"<<"\n";

    return 0;

}

 

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define FRU freopen("out.txt","w",stdout)

#define FRO freopen("in.txt","r",stdin)

#define pb push_back

#define mp make_pair

#define ff first

#define ss second

#define mem(ara,n) memset(ara,n,sizeof ara)

#define loop(i,j,n) for(i=j;i<n;i++)

#define rloop(i,j,n) for(i=n;i>=j;i--)

#define INF 2147483647

#define ll long long

#define pii pair<int,int>

#define eps 1e-9

#define mii map<int,int>

#define vi vector<int>

#define all(n) n.begin(),n.end()

#define inf INF

#define INFLL 9223372036854775807

using namespace std;

int main()

{

    int n,m,cnt=0,i,j,k;

    string s="-",s1;

    int a,b,c,d;

    cin>>n>>a>>b>>c>>d;

    cin>>s1;

    s+=s1;

    int ara[200005]={0};

    if(c>d)

    {

        int flag=0,flag1=0;

        for(i=a;i<=c;i++)

        {

            if(s[i]=='#'&& s[i-1]=='#')flag1=1;

        }

        if(s[b-1]=='.')ara[b-1]=1;

        for(i=b;i<=d;i++)

        {

            if(s[i]=='.')ara[i]=ara[i-1]+1;

            if(ara[i]>=3)flag=1;

        }

        if(s[d+1]=='.'&& ara[d]>1)flag=1;

        if(flag&& !flag1)cout<<"Yes\n";

        else cout<<"No\n";

    }

    else

    {

        int flag=0;

        for(i=a;i<=c;i++)

        {

            if(s[i]=='#'&& s[i-1]=='#')flag=1;

        }

        for(i=b;i<=d;i++)

        {

            if(s[i]=='#'&& s[i-1]=='#')flag=1;

        }

        /*if(c>b&& c<d)

        {

            if(s[c-1]=='#'|| s[c+1]=='#')flag=1;

        }*/

        if(!flag)cout<<"Yes\n";

        else cout<<"No\n";

    }



}


// Optimized solution to the same problem (faster):
#include<cstdio>



using namespace std;



char s[200002];

int main(){

	int n,a,b,c,d;

	scanf("%d%d%d%d%d%s",&n,&a,&b,&c,&d,s+1);

	int r1=a,r2=b;

	for(;;){

		if(r1<n&&s[r1+1]!='#')r1++;

		else if(r1<n-1&&s[r1+2]!='#')r1+=2;

		else if(r2<n&&s[r2+1]!='#')r2++;

		else if(r2<n-1&&s[r2+2]!='#')r2+=2;

		else break;

	}

	if(!(c<=r1)||!(d<=r2)){

		printf("No\n");

		return 0;

	}

	if(c>d){

		bool ok=1;

		for(int i=b-1;i+1<=d;i++)

			if(i+2<=r1&&s[i]!='#'&&s[i+1]!='#'&&s[i+2]!='#')

				ok=0;

		if(ok){

			printf("No\n");

			return 0;

		}

	}

	printf("Yes\n");

	

}

// Fastest known solution for this problem: