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;
	int i;
	int count = 0;

	scanf("%d%d%d%d%d", &N, &A, &B, &C, &D);

	char masu[N];

	for (i = 1;i <= N;i++) {
		scanf("%c", &masu[i]);
	}

	if (C < D) {
		for (i = A;i <= C-1;i++) {
			if (masu[i] == '#'&& masu[i+1]=='#'){
				printf("No\n");
				return 0;
			}
		}

		for (i = B;i <= D - 1;i++) {
			if (masu[i] == '#' && masu[i + 1] == '#') {
				printf("No\n");
				return 0;
			}
		}

		printf("Yes\n");
	}
	else {
		for (i = A;i <= C - 1;i++) {
			if (masu[i] == '#' && masu[i + 1] == '#') {
				printf("No\n");
				return 0;
			}
		}
		for (i = B - 1;i < D - 1;i++) {
			if (masu[i] == '.' && masu[i + 1] == '.' && masu[i + 2] == '.') {
				count++;
			}
		}
		if (count > 0) {
			printf("Yes\n");
		}
		else {
			printf("No\n");
		}
	}

	return 0;

}