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>
#include <string.h>

int main(){
    int N, A, B, C, D, s;
    int flag = 1; // Yes
    int i;
    char c;

    scanf("%d %d %d %d %d", &N, &A, &B, &C, &D);
    getchar();

    A -= 1; B-= 1; C-=1; D -= 1; // 配列の要素番号とそろえる
    int masu[N];
    for(i=0; i<N; i++){
        scanf("%c", &c);
        if(c == '.'){
            masu[i] = 0;
        }else{
            masu[i] = 1;
        }
    }

    //C,Dの前に2つ並んだ石があったら不可能
    for(i=A; (i<N-1 && i < C && i < D); i++){
        if(masu[i] == 1){
            if(masu[i+1] == 1){
                flag = 0;
                break;
            }
        }
    }

    // C > D にも拘わらず　... がないとAがBを抜けない
    if(C > D){
        //printf("point\n");
        flag = 0; 
        for(i=B-1; (i < D-1 && i < N-2); i++){
            if(masu[i] == 0){
                if(masu[i+1] == 0 && masu[i+2] == 0){
                    //printf("%d\n", i);
                    flag = 1;
                }
            }
        }
    }

    //for(i=0; i < N; i++)printf("%d", masu[i]);
    //printf("\n");

    //上記以外はいける　（とおもう）

    if(flag == 1){
        printf("Yes\n");
    }else{
        printf("No\n");
    }
    
    return 0;
}