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>

int main(void){

int n,a,b,c,d, i;
int q = 0;
int z = 0;
char s[200000];
int w[200000];
int p = 0;

scanf("%d", &n);
scanf("%d", &a);
scanf("%d", &b);
scanf("%d", &c);
scanf("%d", &d);
scanf("%s", s);

for(i = 0; i < n; i++){
  if(s[i] == '.') w[i] = 0;
  else w[i] = 1;
}


if(c<d){
  for(i = b+1; i < d; i++){
    if(w[i]*w[i+1] != 0) q = 1;
  }
  for(i = a+1; i < c; i++){
    if(w[i]*w[i+1] != 0) z = 1;
  }
if(q == 0 && z==0) printf("Yes\n");
else printf("No\n");
}

else{
  for(i = b+1; i < d; i++){
    if(w[i]*w[i+1] != 0) q = 1;//先移動
  }
  for(i = a+1; i < c; i++){
    if(w[i]*w[i+1] != 0) z = 1;
  }
  for(i = b+1; i < d-1; i++){
    if(w[i] == 0 && w[i+1] == 0 && w[i + 2] == 0) p = 1;//先移動
  }
if(q == 0 && z == 0 && p == 1) printf("Yes\n");
else printf("No\n");
}


  return 0;
}