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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main()
{
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, a, b, c, d;
  cin >> n >> a >> b >> c >> d;
  --a, --b, --c, --d;
  string t;
  cin >> t;
  auto reachable = [&](int s, int f) {
    if (t[s] == '#' || t[f] == '#') return false;
    for (int i = s + 1; i <= f; ++i) {
      if (t[i] == t[i - 1] && t[i] == '#') {
        return false;
      }
    }
    return true;
  };
  auto hasg3 = [&](int s, int f) {
    for (int i = s; i <= f - 2; ++i) {
      if (t[i] == t[i + 1] && t[i] == t[i + 2] && t[i] == '.') {
        return true;
      }
    }
    return false;
  };
  bool can = true;
  if (!reachable(a, c) || !reachable(b, d)) {
    can = false;
  }
  if (c > d) {
    if (!hasg3(b - 1, d + 1)) {
      can = false;
    }
  }
  cout << (can ? "Yes" : "No") << endl;
  return 0;
}