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 <bits/stdc++.h>
#define int long long
#define ALL(v) (v).begin(),(v).end()
#define Vi vector<int>
#define pb push_back
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repeq(i,a,b) for(int i=a;i<=b;i++)
#define repr(i,a,b) for(int i=a;i>b;i--)
#define repreq(i,a,b) for(int i=a;i>=b;i--)
#define leng(n) (int)(log10(n)+1)
#define INF 9000000000000000000
#define MOD 1000000007
using namespace std;

int gcd(int a,int b){return b?gcd(b,a%b):a;}  //  最大公約数gcd
int lcm(int a,int b){return a/gcd(a,b)*b;}  //  最小公倍数lcm

signed main() {
  int l,as,ag,bs,bg;
  cin >> l >> as >> bs >> ag >> bg;
  string s;
  cin >> s;
  as--;
  ag--;
  bs--;
  bg--;
  
  int amin=0,amax=l-1,bmin=0,bmax=l-1;
  
  rep(i,as,l-1){
    if(s[i]=='#' && s[i+1]=='#'){
      amax = i-1;
      break;
    }
  }
  
  rep(i,bs,l-1){
    if(s[i]=='#' && s[i+1]=='#'){
      bmax = i-1;
      break;
    }
  }
  
  /*repr(i,as,0){
    if(s[i]=='#' && s[i-1]=='#'){
      amin = i+1;
      break;
    }
  }
  
  repr(i,bs,0){
    if(s[i]=='#' && s[i-1]=='#'){
      bmin = i+1;
      break;
    }
  }*/
  
  amin = as;
  bmin = bs;
  
  //cout << amax << ' ' << amin << ' ' << bmax << ' ' << bmin << endl;
  //cout << ag << endl;
  
  if(ag<amin || ag>amax){
    cout << "No" << endl;
    return 0;
  }
  if(bg<bmin || bg>bmax){
    cout << "No" << endl;
    return 0;
  }
  if(ag<bg){
    cout << "Yes" << endl;
    return 0;
  }
  
  bool three = false;
  
  rep(i,amin,amax-1){
    if(s[i]=='.' && s[i+1]=='.' && s[i+2]=='.') three = true;
  }
  
  if(three) cout << "Yes" << endl;
  else cout << "No" << endl;
}