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)

//From Naruto to Hokage
//#include <RJ>
#include<bits/stdc++.h>
using namespace std;

#define ll          long long
#define pb          push_back
#define eb          emplace_back
#define endl        '\n'
#define mii         map<ll,ll>
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define INF         (1ll<<60)
#define rep(i,a,b)  for(ll i=a;i<=b;i++)
#define rrep(i,a,b)  for(ll i=a;i>=b;i--)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }
const int N=100005;
void solve(){
    ll n,a,b,c,d;
    cin>>n>>a>>b>>c>>d;
    a--,b--,c--,d--;
    string s;
    cin>>s;
    ll t=0;
    if(d<c){
        rep(i,b-1,d-1){
            if(s[i]=='.' && s[i+1]=='.' && s[i+2]=='.')
                t=1;
        }
    }
    else t=1;
    if(t==0) {cout<<"No";return;}
    while(1){
        if(a==c) break;
        else if(s[a+2]=='.' && a+2<=c) a+=2;
        else if(s[a+1]=='.' && a+1<=c) a+=1;
        else{cout<<"No";return;}
    }
    while(1){
        if(b==d) break;
        else if(s[b+2]=='.' && b+2<=d) b+=2;
        else if(s[b+1]=='.' && b+1<=d) b+=1;
        else{cout<<"No";return;}
    }
    cout<<"Yes";
}
signed main()
{
    ios
    int TESTS=1;
    //cin>>TESTS;
    while(TESTS--){
        solve();
    }
    time
    return 0;
}