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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//A（敗北）
#include <iostream>
#include <string>
using namespace std;
int Maper(int x);
int maxn = 0;
int can[3];
string S;
 
int main() {
int ans = 0;
int check = 0;
  	
  	int N,A,B,C,D;
  	cin >> N >> A >> B >> C >> D;
  	cin >> S;
  //A-MAX(C,D)間で二連で石がないことの確認
	can[0] = Maper(A);
    can[1] = Maper(A+1);
    can[2] = Maper(A+2);
  	if(can[1] == 0 && ( can[0] * can[2] == 0)){ //絶対渡れない
    	check = 1;
    } else{
	   	for (int i = A+2 + 1;(i<=C || i<=D);i++){ //min(C,D)
        	
			can[0] = can[1];
            can[1] = can[2];
            can[2] = Maper(i);
 
			 if(can[1] == 0 &&( can[0] * can[2] == 0)){ //絶対渡れない
            	check = 1;
              	break;
            }
        }
    }
  if(check == 0){
  //D<Cなら,B-D間で3連で石があるかどうか
  can[0] = Maper(B);
    can[1] = Maper(B+1);
    can[2] = Maper(B+2);
  	if(C<D){ //	絶対渡れる
    	ans = 1;
    } else  if (can[0] * can[1] * can[2] == 1){ //絶対渡れる
    	ans = 1;
    } else
	   	for (int i = B+2 + 1; (i<=C || i<=D) ;i++){ //i<=max(C,D)
        	
			can[0] = can[1];
            can[1] = can[2];
            can[2] = Maper(i);
            if(can[0] * can[1] * can[2] == 1){ //3つとも1なのが存在すれば渡れる
            	ans = 1;
                break;
            }
        }
    }
    if(ans == 1){
    	cout << "Yes";
    } else{
    	cout << "No";
    }
return 0;
  
  
}
int Maper(int x){ //上乗れる？
	if(S.find("#",x-1)+1 == x){
      //cout << "!";
      return 0;
    }
    else{
    	return 1;
    }
}