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 <algorithm>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>

using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
const int INF = (1 << 30);
const ll INFL = (1LL << 62);
const ll MOD = 1000000007;
template <class T>
bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
T gcd(T a, T b) {
    return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
    return a / gcd(a, b) * b;
}
template <class T>
ll mod_pow(ll a, ll n, T mod) {
    mod = (ll)mod;
    ll res = 1, p = a % mod;
    while (n) {
        if (n & 1) res = res * p % mod;
        p = p * p % mod;
        n >>= 1;
    }
    return res;
}
template <int mod>
struct ModInt {
    using ll = long long;
    ll x;
    ModInt() : x(0) {}
    ModInt(int y) : x(y >= 0 ? ll(y % mod) : (ll)((mod - (-y) % mod))) {}
    ModInt(ll y) : x(y >= 0 ? y % mod : (mod - (-y) % mod)) {}
    ModInt &operator+=(const ModInt &rhs) {
        if ((x += rhs.x) >= mod) x -= mod;
        return *this;
    }
    ModInt &operator-=(const ModInt &rhs) {
        if ((x += mod - rhs.x) >= mod) x -= mod;
        return *this;
    }
    ModInt &operator*=(const ModInt &rhs) {
        x = x * rhs.x % mod;
        return *this;
    }
    ModInt &operator/=(const ModInt &rhs) {
        *this *= rhs.inverse();
        return *this;
    }
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &rhs) const { return ModInt(*this) += rhs; }
    ModInt operator-(const ModInt &rhs) const { return ModInt(*this) -= rhs; }
    ModInt operator*(const ModInt &rhs) const { return ModInt(*this) *= rhs; }
    ModInt operator/(const ModInt &rhs) const { return ModInt(*this) /= rhs; }
    ModInt operator==(const ModInt &rhs) const { return x == rhs.x; }
    ModInt operator!=(const ModInt &rhs) const { return x != rhs.x; }
    friend ostream &operator<<(ostream &os, const ModInt &rhs) {
        return os << rhs.x;
    }
    friend istream &operator>>(istream &is, ModInt &rhs) {
        ll t;
        is >> t;
        rhs = ModInt<mod>(t);
        return (is);
    }

   private:
    ModInt inverse() const {
        ll a = x, u = 1LL, v = 0LL, t, b = (ll)mod;
        while (b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }
};
using mint = ModInt<MOD>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, A, B, C, D;
    cin >> N >> A >> B >> C >> D;
    string S;
    cin >> S;
    // if (A == B or C == D) {
    //     cout << "No" << endl;
    //     return 0;
    // }
    if (A < B and C < D) {
        cout << "Yes" << endl;
        return 0;
    } else if (B < A and D < C) {
        cout << "Yes" << endl;
        return 0;
    } else if (B < A and C < D) {
        bool flag = false;
        for (int i = B; i <= C - 1; i++) {
            if (S[i] == '.' and S[i + 1] == '.' and S[i + 2] == '.')
                flag = true;
        }
        if (flag)
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
        return 0;
    } else {
        bool flag = false;
        for (int i = A; i <= D - 1; i++) {
            if (S[i] == '.' and S[i + 1] == '.' and S[i + 2] == '.')
                flag = true;
        }
        if (flag)
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
        return 0;
    }

    return 0;
}