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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W;
double X, Y, Z;

int run(){
    
    int i, j, k, l, m, n;
    char *S;
    
    scanf("%d", &N);
    S = (char *)malloc(sizeof(char)*(N+2));
    scanf("%d", &A);
    scanf("%d", &B);
    scanf("%d", &C);
    scanf("%d", &D);
    scanf("%s", S);
    
    A--;
    B--;
    C--;
    D--;
    
    S[A] = 'A';
    S[B] = 'B';
    
    // printf(":: %s", S);
    
    if(C < D){
        for(i = A; i < C; i++){
            if(S[i+1]!='#') continue;
            if(S[i+1]=='#'){
                i++;
                if(S[i+1]=='#') return 1;
            }
        }
        if(i!=C) return 11;
        for(i = B; i < D; i++){
            if(S[i+1]!='#') continue;
            if(S[i+1]=='#'){
                i++;
                if(S[i+1]=='#') return 2;
            }
        }
        if(i!=D) return 12;
        return 0;
    }else{
        for(i = B; i < D; i++){
            if(S[i+1]!='#') continue;
            if(S[i+1]=='#'){
                i++;
                if(S[i+1]=='#') return 3;
            }
        }
        if(i!=D) return 13;
        int alpha = 0;
        int beta = 0;
        for(i = A; i < C; i++){
            if(S[i+1]!='#'){
                alpha++;
                if(alpha > 2) if(i > D) beta = 1;
                continue;
            }
            if(S[i+1]=='#'){
                alpha = 0;
                i++;
                if(S[i+1]=='#') return 4;
            }
        }
        if(i!=C) return 14;
        if(beta==0){
            if((S[A+1]=='B') && (S[A+2]=='.')) return 0;
            return 5;
        }
        return 0;
    }
}

int main(){
    int ret = run();
    if(ret==0) printf("Yes");
    else printf("No");
    
    // printf("(%d)\n", ret);
    return 0;
}