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 <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define LLINF 1<<63-1
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define MOD 1000000007
#define NUM 100001
void swap(int* x,int* y){
  int tmp=*x;*x=*y;*y=tmp;
}

int main() {
  int N,A,B,C,D,i,ans;
  scanf("%d%d%d%d%d",&N,&A,&B,&C,&D);
  A--;B--;C--;D--;
  char S[N+1];
  scanf("%s",S);
  if(D<C){
    ans=0;
    for(i=A+1;i<C-2;i++)if(S[i]=='.'&&S[i+1]=='.'&&S[i+2]=='.')ans=1;
    for(i=A+1;i<C-1;i++)if(S[i]=='#'&&S[i+1]=='#')ans=0;
  }else if(C<B){
    ans=1;
    for(i=A+1;i<C-1;i++)if(S[i]=='#'&&S[i+1]=='#')ans=0;
    for(i=B+1;i<D-1;i++)if(S[i]=='#'&&S[i+1]=='#')ans=0;
  }else{
    ans=1;
    for(i=A+1;i<D-1;i++)if(S[i]=='#'&&S[i+1]=='#')ans=0;
  }
  if(ans==1)printf("Yes");
  else printf("No");

  //printf("%d\n",ans);
	return 0;
}