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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef pair<int, int>P;
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
typedef vector<vector<ll> > mat;
typedef vector<ll> vec;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
const ll mod = 1000000007;
const ll INF = mod * mod;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//int qp(int a,ll b){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}
ll qp(ll a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} // 最大公約数
ll lcm (ll a, ll b){return a/gcd(a,b)*b;} // 最小公倍数
int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };

template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool chmin(T &m, const T q) { if (q < m) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template<typename T1, typename T2> pair<T1, T2> operator<(const pair<T1, T2> l, const pair<T1, T2> r) { return (l.first < r.first); }
template<typename T1, typename T2> pair<T1, T2> operator>(const pair<T1, T2> l, const pair<T1, T2> r) { return (l.first > r.first); }


void solve() {
    ll N,A,B,C,D; cin >> N >> A >> B >> C >> D;
    A--;B--;C--;D--;
    string S; cin >> S;
    S.push_back('#');
    S.push_back('#');
    string res = "Yes";
    if(C<D){
        for(int i=A; i<D;){
            if(S[i+1]=='.')i++;
            else if(S[i+2]=='.'){
                i+=2;
                if(i>D){
                    res = "No";
                    break;
                }
            }
            else{
                res = "No";
                break;
            }
        }
    }
    else{
        bool ok = false;
        for(int i=A; i<C;){
            if(S[i+1]=='.'){
                if(i>=B && i<=D && S[i-1]=='.') ok = true;
                i++;
            }
            else if(S[i+2]=='.'){
                i+=2;
                if(i>C){
                    res = "No";
                    break;
                }
            }
            else{
                res = "No";
                break;
            }
        }
        if(!ok) res = "No";
    }

    cout << res << endl;

}

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

    std::cout << fixed << setprecision(10);
    solve();
    return 0;
}