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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define eee(a) (cerr << (#a) << ": " << (a) << endl);


using ll = long long;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
using mint = map<int, int>;

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

const char sp = ' ';
const char cmm = ',';
const int MOD = 1e9+7;
const int INF = 1e9;
const ll LINF = 1e18;

ll mod(ll a, ll b){return (a%b+b)%b;}
ll gcd(ll a, ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b){return a*b/gcd(a,b);}
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void Judge(bool b){b?Yes():No();}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void JUDGE(bool b){b?YES():NO();}
ll powMod(ll b, ll e, ll m){ll r=1;while(e>0){if(e&1)r=(r%m)*(b%m)%m;b=(b%m)*(b%m)%m;e>>=1;}return r;}
double distance(ll x1, ll y1, ll x2, ll y2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T>
void ppp(T n){cout << n << endl;}
template<typename T1, typename T2> ostream& operator<<(ostream& s, const pair<T1, T2>& p) {return s << "(" << p.first << ", " << p.second << ")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& v) {
    int len = v.size();
    s << '[';
    for (int i = 0; i < len; ++i) {
        s << v[i]; if (i < len-1) s << ", ";
    }
    s << ']';
    return s;
}

struct aaa{
    aaa(){
        cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(15);
    };
}fastio;

int main(){

    int n, a, b, c, d;
    cin >> n >> a >> b >> c >> d;
    --a; --b; --c; --d;

    string s;
    cin >> s;

    if(c < d){
        for(int i = b+1; i < d-1; ++i){
            if(s[i] == '#' and s[i+1] == '#'){
                No();
                return 0;
            }
        }
        for(int i = a+1; i < c-1; ++i){
            if(s[i] == '#' and s[i+1] == '#'){
                No();
                return 0;
            }
        }
    }

    else{
        int idx = -1;
        for(int i = b-1; i < d; ++i){
            if(s.substr(i, 3) == "...") idx = i+1;
        }
        if(idx >= 0){
            s[idx] = '#';
            for(int i = a+1; i < c-1; ++i){
                if(s[i] == '#' and s[i+1] == '#'){
                    No();
                    return 0;
                }
            }
            for(int i = b+1; i < d-1; ++i){
                if(s[i] == '#' and s[i+1] == '#'){
                    No();
                    return 0;
                }
            }
        }
        else{
            s[b] = '#';
            for(int i = a+1; i < c-1; ++i){
                if(s[i] == '#' and s[i+1] == '#'){
                    No();
                    return 0;
                }
            }
            s[b] = '.';
            for(int i = b+1; i < d-1; ++i){
                if(s[i] == '#' and s[i+1] == '#'){
                    No();
                    return 0;
                }
            }
        }
    }
    Yes();
    return 0;
}