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)

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fu(i,r,t) for(int i=r;i<=t;i++)
#define fd(i,r,t) for(int i=r;i>=t;i--)
#define Set set<int>
#define int long long
#define P pair<int,int>
const int maxn = 1e6+5;
const int INF = 1e17 +6;
using namespace std;
//int n,m,a[maxn],b[maxn],ans;
int n;
void over()
{
    cout<<"No";
    exit(0);
}
int a,b,c,d,len;
string k;
bool if_have(int num,int kinds,int st,int zhong)
{
    //kinds:0-#,1-. ,包括st
    int times=0;
    if(kinds)
    {
        fu(i,st,zhong)
        {
            if(k[i]=='.')
            {
                times++;
                if(times==num)return 1;
            }else{
                times=0;
            }
        }
    }else{
        fu(i,st,zhong)
        {
            if(k[i]=='#')
            {
                times++;
                if(times==num)return 1;
            }else{
                times=0;
            }
        }
    }
    return 0;
}
int32_t main()
{
    IOS;
    cin>>n>>a>>b>>c>>d;
    cin>>k;
    len=k.length();
    int time1=0;
    if(c<d)//b
    {
        if(if_have(2,0,b-1,d-1))over();
        if(if_have(2,0,a-1,c-1))over();
        cout<<"Yes";
    }else{
        //看看能不能先走a

        //看看b能不能挪
        if(if_have(3,1,b-2,d-1)){//可以挪
            if(if_have(2,0,a-1,c-1))over();
            if(if_have(2,0,b-1,d-1))over();
            cout<<"Yes";
        }else{
            if(if_have(2,0,b-1,d-1))over();
            k[d-1]='#';
            if(if_have(2,0,a-1,c-1))over();
            cout<<"Yes";
        }
    }
    return 0;
}