Problem:
Score : 400 points

Problem Statement

There are N squares arranged in a row, numbered 1, 2, ..., N from left to right.
You are given a string S of length N consisting of . and #. If the i-th character of S is #, Square i contains a rock; if the i-th character of S is ., Square i is empty.

In the beginning, Snuke stands on Square A, and Fnuke stands on Square B.

You can repeat the following operation any number of times:

Choose Snuke or Fnuke, and make him jump one or two squares to the right. The destination must be one of the squares, and it must not contain a rock or the other person.

You want to repeat this operation so that Snuke will stand on Square C and Fnuke will stand on Square D.

Determine whether this is possible.

Constraints

4 \leq N \leq 200\ 000

S is a string of length N consisting of . and #.

1 \leq A, B, C, D \leq N

Square A, B, C and D do not contain a rock.

A, B, C and D are all different.

A < B

A < C

B < D

Input

Input is given from Standard Input in the following format:

N A B C D
S

Output

Print Yes if the objective is achievable, and No if it is not.

Sample Input 1

7 1 3 6 7
.#..#..

Sample Output 1

Yes

The objective is achievable by, for example, moving the two persons as follows. (A and B represent Snuke and Fnuke, respectively.)

A#B.#..

A#.B#..

.#AB#..

.#A.#B.

.#.A#B.

.#.A#.B

.#..#AB

Sample Input 2

7 1 3 7 6
.#..#..

Sample Output 2

No

Sample Input 3

15 1 3 15 13
...#.#...#.#...

Sample Output 3

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//By TheOneYouWant

#pragma GCC optimize ("-O2")
#include <bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define memreset(a) memset(a,0,sizeof(a))
#define testcase(t) int t;cin>>t;while(t--)
#define forstl(i,v) for(auto &i: v)
#define forn(i,e) for(int i = 0; i < e;++i)
#define forsn(i,s,e) for(int i = s; i < e;++i)
#define rforn(i,s) for(int i = s; i >= 0;--i)
#define rforsn(i,s,e) for(int i = s; i >= e;--i)
#define bitcount(a) __builtin_popcount(a) // set bits (add ll)
#define ln '\n'
#define getcurrtime ((double)clock()/CLOCKS_PER_SEC)
#define dbgarr(v,s,e) cerr<<#v<<" = "; forsn(i,s,e) cerr<<v[i]<<", "; cerr<<endl
#define inputfile freopen("input.txt", "r", stdin)
#define outputfile freopen("output.txt", "w", stdout)
#define dbg(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); \
stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) { cerr<<endl; }
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << "\t"; err(++it, args...);
}
template<typename T1,typename T2>
ostream& operator <<(ostream& c,pair<T1,T2> &v){
    c<<"("<<v.fi<<","<<v.se<<")"; return c;
}
template <template <class...> class TT, class ...T>
ostream& operator<<(ostream& out,TT<T...>& c){
    out<<"{ ";
    forstl(x,c) out<<x<<" ";
    out<<"}"; return out;
}

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll,ll> p64;
typedef pair<int,int> p32;
typedef pair<int,p32> p96;
typedef vector<ll> v64;
typedef vector<int> v32; 
typedef vector<v32> vv32;
typedef vector<v64> vv64;
typedef vector<p32> vp32;
typedef vector<p64> vp64;
typedef vector<vp32> vvp32;
typedef map<int,int> m32;

const int LIM=1e5+5, MOD=1e9+7;
const ld EPS = 1e-9;
// int t,n,m,k,x,y,w;
signed main(){
    fastio;
    
    int n,a,b,c,d;
    string s;
    cin>>n>>a>>b>>c>>d;
    cin>>s;

    if(d>c){
        bool p = 1;
        forsn(i,b-1,d){
            if(s[i] == '#'){
                if(i+1 != n && s[i+1] == '#') p = 0;
            }
        }
        forsn(i,a-1,c){
            if(s[i] == '#'){
                if(i+1 != n && s[i+1] == '#') p = 0;
            }
        }
        if(!p) cout<<"No"<<ln;
        else cout<<"Yes"<<ln;
    }
    else if(c>d){
        bool p1 = 1;
        forsn(i,b-1,d){
            if(s[i] == '#'){
                if(i+1 != n && s[i+1] == '#') p1 = 0;
            }
        }
        forsn(i,a-1,c){
            if(s[i] == '#'){
                if(i+1 != n && s[i+1] == '#') p1 = 0;
            }
        }
        bool p2 = 0;
        forsn(i,b-1,d){
            if(i >= 1 && s[i-1] == '.'){
                if(s[i] == '.'){
                    if(i+1 != n && s[i+1] == '.') p2 = 1;
                }
            }
        }
        if(p1 & p2) cout<<"Yes"<<ln;
        else cout<<"No"<<ln;
    }

    return 0;
}