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 < (n); ++i)
#define repr(i, n) for(int i = (n-1); i >= 0; --i)

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 200005;
const int INF = 1001001001;
const int MOD = 1000000007;

int main(){
    int N, A, B, C, D;
    string S;
    cin >> N >> A >> B >> C >> D >> S;
    A--, B--, C--, D--;

    bool res = false;
    if (C > D) {
        for (int i = A; i < D; ++i) {
            if (i+2 < N && S[i] == '.' && S[i+1] == '.' && S[i+2] == '.') {
                res = true;
            }
        }
    } else {
        res = true;
    }

    while (B < D) {
        if (B+1 <= D && S[B+1] == '.') B++;
        else if (B+2 <= D && S[B+2] == '.') B += 2;
        else {
            res = false;
            break;
        }
    }

    while (A < C) {
        if (A+1 <= C && S[A+1] == '.') A++;
        else if (A+2 <= C && S[A+2] == '.') A += 2;
        else {
            res = false;
            break;
        }
    }

    cout << (res ? "Yes" : "No") << endl;
}