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>
#define SORT(a) sort(a.begin(), a.end())
#define RSORT(a) sort(a.rbegin(), a.rend())
#define REP(i, n) for (int i = 0; i < n; i++)
#define RREP(i, n) for (int i = n - 1; 0 <= i; i--)
#define FOR(i, start, end) for (int i = start; i < end; i++)
#define RFOR(i, start, end) for (int i = start - 1; 0 <= i; i--)
#define ALL(a) a.begin(), a.end()
using ll = long long;
using namespace std;
constexpr int INF32       = 1'050'000'000;
constexpr long long INF64 = 4'000'000'000'000'000'000;
constexpr int MOD7        = 1'000'000'007;
constexpr int MOD9        = 1'000'000'009;
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
void print() { std::cout << '\n'; }
template <class H, class... T> void print(H &&head, T &&... args) {
    std::cout << head;
    sizeof...(args) == 0 ? std::cout << "" : std::cout << ' ';
    print(std::forward<T>(args)...);
}
template <class T> void print(std::vector<T> &v) {
    for (int i = 0; i < v.size(); i++) {
        std::cout << v[i];
        i == v.size() - 1 ? std::cout << '\n' : std::cout << ' ';
    }
}
template <class T> void print(std::vector<std::vector<T>> &v) {
    for (int i = 0; i < v.size(); i++) {
        for (int j = 0; j < v[i].size(); j++) {
            std::cout << v[i][j];
            j == v[i].size() - 1 ? std::cout << '\n' : std::cout << ' ';
        }
    }
}
void debug() { std::cerr << '\n'; }
template <class H, class... T> void debug(H &&head, T &&... args) {
    std::cerr << head;
    sizeof...(args) == 0 ? std::cerr << "" : std::cerr << ' ';
    debug(std::forward<T>(args)...);
}
template <class T> void debug(std::vector<T> &v) {
    for (int i = 0; i < v.size(); i++) {
        std::cerr << v[i];
        i == v.size() - 1 ? std::cerr << '\n' : std::cerr << ' ';
    }
}
template <class T> void debug(std::vector<std::vector<T>> &v) {
    for (int i = 0; i < v.size(); i++) {
        for (int j = 0; j < v[i].size(); j++) {
            std::cerr << v[i][j];
            j == v[i].size() - 1 ? std::cerr << '\n' : std::cerr << ' ';
        }
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, A, B, C, D;
    cin >> N >> A >> B >> C >> D;
    A--;
    B--;
    C--;
    D--;
    string s;
    cin >> s;
    auto yes = []() {
        print("Yes");
        exit(0);
    };
    auto no = []() {
        print("No");
        exit(0);
    };

    FOR(i, 1, N) {
        if (s[i - 1] == '#' && s[i] == '#' &&
            ((A < i && i < C) || (B < i && i < D)))
            no();
    }
    if (D < C) {
        FOR(i, B, D) {
            if (s[i - 1] == '.' && s[i] == '.' && s[i + 1] == '.') {
                yes();
            }
        }
        no();
    }
    yes();

    return 0;
}