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;

//vector…可変長、データの大きさに合わせてサイズを変える
//list…双方向リスト、要素を挿入、削除の多いとき
//forward_list
//array…固定長配列
//set…集合
//multiset
//bitset…2進数、固定長ビット列
//map…連想配列、キーと値を関連付ける
//multimap
//deque
//stack…FILO
//queue…FIFO
//priority_queue…格納された最大値から順に取り出す
//pair…二つの値を一組
//tuple…複数の値を一組

template<typename T>
void remove(std::vector<T>& vector, char index) {
    vector.erase(vector.begin() + index);
}

template<class T>
T ma(std::vector<T> &v) {
    T maxi = 0;
    for(auto &i:v) maxi = max(maxi, i);
    return maxi;
}

template<class T>
T mi(std::vector<T> &v) {
    T mini = 0;
    for(auto &i:v) mini = min(mini, i);
    return mini;
}

template <class T>
T pow(T x, T y, T n = numeric_limits<T>::max()) {
    if(y == 0) return 1;
    if(y % 2 == 0) return pow((x * x) % n, y / 2);
    return (x * pow(x, y - 1)) % n;
}

#define rep(i, n, m) for (int i = n; i <= m; i++)
#define Rep(i, n, m) for (int i = m; i >= 1; i--)

typedef long long ll;
typedef unsigned long long ull;

ull ans = 0;
/*
int gcd(int x, int y) {
    if (x < y) swap(x, y);
    return !y ? x : gcd(x % y, y);
}
*/
int main() {
    ull n, a, b, c, d;
    cin >> n >> a >> b >> c >> d;

    string s;
    cin >> s;

    rep(i, b-1, max(c, d)-1) {
        if (s[i] == '#' && s[i+1] == '#') {
            cout << "No" << endl;
            return 0;
        }
    }

    if (c < d) {
        cout << "Yes" << endl;
        return 0;
    }

    rep(i, b, min(c, d)-2) {
        if (s[i] == '.' && s[i+1] == '.' && s[i+2] == '.') {
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
}