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: Wrong Answer)

#include <bits/stdc++.h>
#include <time.h>
using namespace std;
#define ll          long long int
#define pb          push_back
#define pll         pair<ll,ll>
#define speed_up    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mod         1000000007
#define inf         1000000000000000000LL
#define vl          vector<ll>
#define F           first
#define S           second
#define ld          long double
#define sz(x)       (ll)x.size()
#define fix(n)      cout<<fixed<<setprecision(n)
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define all(x)      x.begin(),x.end()
#define pi          3.14159265358979323

ll n, a, b, c, d;
string s;

bool isp(ll x, ll y)
{
       for(ll i=x; i+2<y; i++)
       {
              if(s[i+1]=='#' && s[i+2]=='#')
                     return 0;
       }
       return 1;
}

void solve()
{
     cin>>n>>a>>b>>c>>d>>s;
     a--, b--, c--, d--;
     if(d>c)
     {
            if(isp(a,c) && isp(b,d)) { cout<<"Yes\n"; return; }
            else { cout<<"No\n"; return; }
     }
     else
     {
            s[b]='#';
            if(isp(a,c) && isp(b,d))
            {
                   cout<<"Yes\n";
                   return;
            }
            s[b]='.';
            if(isp(b,d))
            {
                   s[b]='.';
                   s[d]='#';
                   if(isp(a,c))
                   {
                          cout<<"Yes\n";
                          return;
                   }
            }
            ll cd=0, tmp=0;
            s[b]='.';
            s[d]='.';
            for(ll i=b; i<=d; i++)
            {
                   if(s[i]=='.') tmp++;
                   else if(s[i]=='#') { cd=max(cd,tmp); tmp=0;}
            }
            s[b]='.', s[d]='.';
            if(cd>=3 && isp(b,d) && isp(a,c))
            {
                   cout<<"Yes\n";
                   return;
            }
     }
     cout<<"No\n";
}

int main()
{
       speed_up;
       //pre();
       ll T=1;
       //cin>>T;
       while(T--) solve();
	return 0;
}