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 <stdbool.h>
#define ll long long int
#define lim 100010
#define mod 1e9+7
#define INF 1e9
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define ABS(x) ((x)>0?(x):-(x))
int compare(const void *a,const void *b){return *(int*)a-*(int*)b;}

int main(void){
	int N,A,B,C,D;
	scanf("%d%d%d%d%d",&N,&A,&B,&C,&D);
	char S[200001];
	scanf("%s",S);
	bool ans = true;
	for(int i=0;i<strlen(S)-1;i++){
		if(S[i]=='#' && S[i+1]=='#'){
			ans = false;
			break;
		}
	}
	if(C>D && ans){
		ans = false;
		B--;D--;
		for(int i=B;i<=D;i++){
			if(S[i]=='.' && S[i+1]=='.' && S[i-1]=='.'){
				ans = true;
				break;
			}
		}
	}
	
	
	printf("%s",ans?"Yes":"No");
	return 0;
}