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
// ------------------------------------------------
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

// func
// ------------------------------------------------
int CalcSumOfDigit(int n);      // 各桁の和を計算する。
long long getDigit(long long n);      // 数字の桁数を取得する。
string upper(string str);       // 英字を大文字に変換する。
string lower(string str);       // 英字を小文字に変換する。

// define
// ------------------------------------------------
#define all(a)        (a).begin(),(a).end()
#define rall(a)       (a).rbegin(),(a).rend()
#define sz(a)         int((a).size())
#define rep(i,n)      for(int(i)=0;(i)<(n);(i)++)
#define repe(i,n)     for(int(i)=0;(i)<=(n);(i)++)
#define vsort(v)      sort((v).begin(),(v).end())
#define rvsort(v)     sort(rall((v)))
#define vi            vector<int>
#define GCD(a,b)      __gcd((a),(b))
#define LCM(a,b)      ((a)/GCD((a),(b))*(b))
#define kiriage(a,b)  ((a)+(b)-1)/(b)

const int INF = 1e9;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vll;

// code
// ------------------------------------------------
int main()
{
  ll n,a,b,c,d;
  cin >> n >> a >> b >> c >> d;
  string s;cin >> s;

  ll cnt = 0;
  ll blocnt = 0;

  bool flg_3 = false;
  bool flg_blo = false;
  set<ll> pos;
  rep(i,sz(s))
  {
    if(s[i] == '.')
    {
      cnt++;
      blocnt = 0;
    }
    else
    {
      cnt = 0;
      blocnt++;
    }
    
    if(cnt >= 3)
    {
      flg_3 = true;
      pos.insert(i);
    }
    if(blocnt >= 2)
    {
      flg_blo = true;
    }
  }

  if(flg_blo)
  {
    cout << "No" << endl;    
  }
  else
  {
    if(c < d)
    {
      cout << "Yes" << endl;
    }
    else
    {
      bool flg = false;
      for(auto it:pos)
      {
        if(b < it && it < d)
        {
          flg = true;
        }
      }

      if(flg)
      {
        cout << "Yes" << endl;
      }
      else
      {
        cout << "No" << endl;
      }
    }
  }
  



  return 0;
}
// funcの実体
// ------------------------------------------------
ll getDigit(ll n)
{
  int i = 1;
  while(1)
  {
    n = n / 10;
    if(n == 0)
      break;
    i++;
  }
  return i;
}
ll CalcSumOfDigit(ll n)
{
  ll s = 0;
  while(n)
  {
    s += n % 10;
    n = n / 10;
  }
  return s;
}

string upper(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(97 <= *itr && *itr <= 122)
    {
      *itr = *itr - 32;
    }
  }
  return str;
}

string lower(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(65 <= *itr && *itr <= 90)
    {
      *itr = *itr + 32;
    }
  }
  return str;
}