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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include<stdio.h>
#include<string.h>


//文字列Sの中に#が2つ連続で入っているかを調べる
//入っているなら1,入っていないなら0を返す
int check_sharp(int N, char S[N])
{
    int count = 0;    
    for(int i = 0;i < N; i++){
        if(S[i] == '#'){
            count = count + 1;
            if(count >= 2){
                return 1;
            }
        }
        else{
            count = 0;
        }
    }
    return 0;
    
                    
}

//文字列Sの中に.が３つ以上並んでいるところがあるかを調べる
//あるなら0, ないなら１を返す
int check_indent(int N, char S[N])
{
    int count = 0;    
    for(int i = 0;i < N; i++){
        if(S[i] == '.'){
            count = count + 1;
            if(count >= 3){
                return 0;
            }
        }
        else{
            count = 0;
        }
    }
    return 1;
}


int main()
{
    int N, A, B, C, D;
    char S[N];
    int flag1 = 0;
    int flag2 = 0;
    
    scanf("%d %d %d %d %d", &N, &A, &B, &C, &D);
    scanf("%s", S);
    if(C > D){
        flag1 = check_indent(N, S);
        flag2 = check_sharp(N, S);
        if(flag1 == 0 && flag2 == 0){
            printf("Yes");
        }
        else{
            printf("No");
        }
    }
    else{
        if(check_sharp(N, S) == 0){
            printf("Yes");
        }
        else{
            printf("No");
        }
    }

    return 0;
}