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)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 100100100;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
struct Edge {
  ll to;
  ll cost;
};
int main() {
  int N, A, B, C, D;
  cin >> N >> A >> B >> C >> D, A--, B--, C--, D--;
  string S;
  cin >> S;
  for (int i = A; i <= max(C, D) - 1; i++) {
    if (S[i] == '#' && S[i + 1] == '#') {
      cout << "No" << endl;
      return 0;
    }
  }
  if (D < C) {
    bool t = false;
    for (int i = B - 1; i <= D - 2; i++) {
      if (S[i] == '.' && S[i + 1] == '.' && S[i + 2] == '.') t = true;
    }
    if (t) {
      cout << "Yes" << endl;
    } else {
      cout << "No" << endl;
    }
    return 0;
  } else {
    cout << "Yes" << endl;
  }
}