// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



typedef long long               ll;

typedef long double             ld;



#define min3(a, b, c)           min((a), min((b), (c)))

#define max3(a, b, c)           max((a), max((b), (c)))

#define lcm(a,b)                (a / __gcd(a,b) ) *b

#define gcd(a,b)                __gcd(a,b)

#define dist(x1,y1,x2,y2)       sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1))

#define distl(x1,y1,x2,y2)      sqrtl((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1))

#define sqr(a)                  ((a)*(a))

#define urs(r...)               typename decay<decltype(r)>::type

#define FOR(i,a,b)              for(urs(b) i=a;i<b;i++)

#define ROF(i,a,b)              for(urs(b) i=a;i>=b;i--)

#define REP(a,b)                for(urs(b) i=a;i<b;i++)

#define pb                      push_back

#define mk                      make_pair

#define ff                      first

#define ss                      second

#define lb                      lower_bound

#define ub                      upper_bound

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

#define vi                      vector <int> 

#define pi                      pair <int, int> 

#define SORT(v)                 sort(v.begin(), v.end());

#define FASTIO                  ios_base::sync_with_stdio(0);cin.tie(NULL);

#define PI                      acos(-1.0)

const int MOD = 1e9 + 7;



ll fast_pow (ll n, ll p)

{

    ll res=1;

    while (p>0)

    {

        if (p&1)

            res = res * n;

        p = p>>1;

        n = n * n;

    }

    return res;

}



ll BMod(ll B,ll P,ll M)

{

    ll R=1;

    B=B%M;

    while(P)

    {

        if(P&1)

            R=(R*B)%M;

        P >>= 1;

        B=(B*B)%M;

    }

    return R;

}



ll invMOD(ll num,ll M)

{

    return BMod(num,M-2,M);

}



#ifdef Sieve

const int pSz = 50005;

bool np[pSz + 10]; vi prime; int prime_size;

void sieve(){np[0] = np[1] = 1;prime.pb(2);

for(ll i = 4; i <= pSz; i+=2) np[i] = 1;

for(ll i = 3; i <= pSz; i+=2){if(!np[i]){prime.pb(i);

for(ll j = i * i; j <= pSz; j += (i << 1)) np[j] = 1;}}prime_size = prime.size();}

#endif



#ifdef Combi

const int nSz = 2000006;

ll F[nSz + 1], tMod = MOD;

void Factorial(){ F[0] = 1; for(int i = 1; i <= nSz; i++) F[i] = (F[i - 1] * i) % tMod; }

inline ll nCr(int n, int r) { if(n < r) return 0; return (F[n] * invMOD((F[n - r] * F[r]) % tMod, tMod)) % tMod; }

 

#endif

 

#ifdef Z_Algo

void zAlgo(char *s, int *z){

    int L, R, sz; sz = strlen(s); z[0] = L = R = 0;

    FOR(i, 1, sz) { z[i] = 0; if(i <= R) z[i] = min(z[i - L], R - i + 1); 

    while(i + z[i] < sz && s[i + z[i]] == s[z[i]]) z[i]++; if(i + z[i] - 1 > R) L = i, R = i + z[i] - 1; }

}void zAlgo(string &s, int *z){

    int L, R, sz; sz = s.size(); z[0] = L = R = 0;

    FOR(i, 1, sz) { z[i] = 0; if(i <= R) z[i] = min(z[i - L], R - i + 1); 

    while(i + z[i] < sz && s[i + z[i]] == s[z[i]]) z[i]++; if(i + z[i] - 1 > R) L = i, R = i + z[i] - 1; }

}void zAlgo(int *s, int *z, int n){

    int L, R, sz; sz = n; z[0] = L = R = 0;

    FOR(i, 1, sz) { z[i] = 0; if(i <= R) z[i] = min(z[i - L], R - i + 1); 

    while(i + z[i] < sz && s[i + z[i]] == s[z[i]]) z[i]++; if(i + z[i] - 1 > R) L = i, R = i + z[i] - 1; }

}

#endif





//////////////////////////////////////////////////////////////////////////////

///                           M. K. BASHAR                                 ///

///                            IUT CSE'18                                  ///

//////////////////////////////////////////////////////////////////////////////



int main()

{



    unsigned n, a, b, c, d;

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

    string s;

    cin >> s;



    a--;

    b--;

    c--;

    d--;

    bool ck1, ck2, ck3;

    

    if (s.find ("##", b) < d)

        ck1 = true;

    else

        ck1 = false;



    if (s.find ("##", a) < c)

        ck2 = true;

    else

        ck2 = false;



    if (c<d)

    {

        if (ck1 || ck2)

            cout << "No";

        else

            cout << "Yes";

    }

    else

    {



        if (s.find("...", b-1) <= d-1)

            ck3 = false;

        else

            ck3 = true;



        if (ck1 || ck2 || ck3)

            cout << "No";

        else

            cout << "Yes";

    }



    return 0;

}


// 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: