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>
#define min(a,b) ((a)<(b)?(a):(b))

int main(){
	int n,a,b,c,d,i;
	char s[200101];
	scanf("%d%d%d%d%d%s",&n,&a,&b,&c,&d,s);
	for(i=1;i<n;i++){
		if(s[i-1]=='#'&&s[i]=='#'){
			puts("No");
			return 0;
		}
	}
	a--;
	b--;
	c--;
	d--;
	int ans=0;
	int dp[200011]={0},ddp[200011]={0};
	dp[b]=1;
	int f=0;
	for(i=b;i<=d;i++){
		if(dp[i]>0){
		if(s[i+1]!='#')dp[i+1]++;
		if(s[i+2]!='#')dp[i+2]++;
		}
	}
	if(dp[d]==0){
		puts("No");
		return 0;
	}ddp[a]=1;
	for(i=a;i<=c;i++){
		if(ddp[i]>0){
		if(s[i+1]!='#')ddp[i+1]++;
		if(s[i+2]!='#')ddp[i+2]++;
		}
	}
	if(dp[c]==0){
		puts("No");
		return 0;
	}
	if(d>c){
		puts("Yes");
		return 0;
	}
	for(i=b;i<=d;i++){
		if(ddp[i]>0&&dp[i]>0&&(dp[i]==1||ddp[i]==1)&&f!=1){
			f=-1;
		}
		if(dp[i]==2&&ddp[i]==2)f=1;
	}
	if(f==1)puts("Yes");
	else puts("No");
	return 0;
	/*
	for(i=b;i<=d;i++){
		if(dp[i]==2)f=1;
	}
	if(f==0)s[d]='#';
	ddp[a]=1;
	for(i=a;i<=c;i++){
		if(ddp[i]>0){
		if(s[i+1]!='#')ddp[i+1]++;
		if(s[i+2]!='#')ddp[i+2]++;
		}
	}
	if(ddp[c]==0){
		if(f==1){
			puts("No");
			return 0;
		}
		else f=3;
	}
	else {
		puts("Yes");
		return 0;
	}
	for(i=0;i<n;i++){
		dp[i]=0;
		ddp[i]=0;
	}
	s[d]='.';
	ddp[a]=1;
	for(i=a;i<=c;i++){
		if(ddp[i]>0){
		if(s[i+1]!='#')ddp[i+1]++;
		if(s[i+2]!='#')ddp[i+2]++;
		}
	}
	if(ddp[c]==0){
		puts("No");
		return 0;
	}
	for(i=a;i<=c;i++){
		if(ddp[i]==2)f=1;
	}
	if(f==3)s[c]='#';
	dp[b]=1;
	for(i=b;i<=d;i++){
		if(dp[i]>0){
		if(s[i+1]!='#')dp[i+1]++;
		if(s[i+2]!='#')dp[i+2]++;
		}
	}
	if(dp[d]==0)puts("No");
	else puts("Yes");*/
	return 0;
}