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)

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// C++ Conpetitive Programming Template 
// Ver 1.1 (codename: Hinata)
// Last Modified : 2019-05-28
// Created       : 2019-05-25
// Auther        : shirataki3
#include <bits/stdc++.h>
using namespace std;
// typename alias
template <class T> using vec = vector<T>;
template <class T, class U> using vep = vec<pair<T,U> >;
template <class T> using PQ = priority_queue<T>;
using ll = long long;
using vll = vec<ll>;
using pll = pair<ll, ll>;
using vpll = vec<pll>;
using str = string;
// あくまのちから
//#define int ll

// const
static const long long INF = 1LL<<59;
static const long long MOD = (ll)1e9+7;
static const double PI = 3.14159265358979323846;

// accelerate stdin, stdout
#define endl "\n"
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
// repeat macro
#define FOR(i, a, b) for(ll (i)=(a);(i)<=(b);i++)
#define rFOR(i, a, b) for(ll (i)=(b);(i)>=(a);i--)
#define REP(i, n) for(ll i=0;i<n;i++)
// vector macro
#define WHOLE(v) (v).begin(), (v).end()
#define ASC(v) sort(WHOLE(v))
#define DESC(v) sort((v).rbegin(), (v).rend())
#define INV(v) reverse(WHOLE(v))
#define FOLDL(src, dst, lambda) partial_sum(WHOLE(src), (dst).begin(), lambda)
#define FOLDR(src, dst, lambda) INV(src);partial_sum(WHOLE(src), (dst).begin(), lambda);INV(dst)
#define CUMSUML(src, dst) FOLDL(src, dst, [](auto a, auto b) {return a+b;})
#define CUMSUMR(src, dst) FOLDR(src, dst, [](auto a, auto b) {return a+b;})
#define ACCUM(v, d0, lambda) accumulate(WHOLE(v), d0, lambda)
#define SUM(v, d0) ACCUM(v, d0, [](auto a, auto b) {return a+b;})
#define ERASE(v, i) v.erase(v.begin() + i)
#define UNIQ(v) ASC(v);v.erase(unique(v.begin(), v.end()), v.end())

// io helper
template <typename T>void debug(T t) {cerr << t << endl;}
template <typename T, typename ...Args>void debug(T t, Args ...args) {cerr << t << " ";debug(args...);}
template <typename T>void out(T t) {cout << t << endl;}
template <typename T, typename ...Args>void out(T t, Args ...args) {cout << t << " ";out(args...);}
template <typename T>istream& operator>>(istream &is, vec<T> &v) {for(auto &a:v) is >> a; return is;}
template <typename T>ostream& operator<<(ostream &os, vec<T> &v) {os << "[";REP(i, v.size() - 1) os << v[i] <<", "; os << v[v.size()-1]<<"]";return os;}
template <typename T, typename U>ostream& operator<<(ostream &os, pair<T, U> &p) {os << "(" << p.first << ", " << p.second << ")";return os;}
template <typename T, typename U>ostream& operator<<(ostream &os, map<T, U> &mp) {os << "{ ";for(auto &p:mp) os << p.first << ": " << p.second << ", "; os << "}";return os;}

void solve() {
    ll N, A, B, C, D;
    str S;
    cin >> N >> A >> B >> C >> D;
    cin >> S;
    replace(S.begin(), S.end(), '#', 'X');
    replace(S.begin(), S.end(), '.', 'O');
    // a-b-c-d
    if(C < D) {
        const str atod = S.substr(A, D-A);
        smatch m;
        auto f = regex_search(atod, m, regex("XX"));
        if(f) {
            out("No");
        }else {
            out("Yes");
        }
    }else {
    // a-b-d-c
        const str atoc = S.substr(A, C-A);
        smatch m;
        bool f = regex_search(atoc, m, regex("XX"));
        if(f) {
            out("No");
            return;
        }
        f = regex_search(atoc, m, regex("OOO"));
        if(f) {
            out("Yes");
            return;
        }
        out("No");
    }
}

signed main() {
    fast_io;
    cout << setprecision(20) << fixed;
    solve();
    return 0;
}