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)

/*
ID: thanhvq1
LANG: C++14
PROB: namenum
*/

#pragma GCC optimize ("O2")
#include<bits/stdc++.h>
#include<unistd.h>
using namespace std;
//#define int long long
#define double long double
#define vec vector
#define vi vector<int>
#define FZ(n) memset((n),0,sizeof(n))
#define FMO(n) memset((n),-1,sizeof(n))
#define unset unordered_set
#define ii pair<int, int>
#define vii vector<ii>
#define F first
#define S second
#define PB push_back
#define ALL(x) begin(x),end(x)
#define SZ(x) ((int)(x).size())
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define FORE(i, a, b) for(int i = (a); i <= (b); i++)
#define FORD(i, a, b) for(int i = (a); i >= (b); i--)
#define bitcount(n) __builtin_popcountll(n)
template <typename T,typename U>
std::pair<T,U> operator+(const std::pair<T,U> & l,const std::pair<T,U> & r) {
  return {l.first+r.first, l.second+r.second};
} 
typedef void (*callback_function)(void); // type for conciseness
const long long ZERO = 0LL;
const long long INF64 = 1e18;
const int INF32 = 1e9;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0L);
const double EPS = static_cast<double>(1e-9);
typedef long long ll;

void FILEIO() {  freopen("in.txt", "r", stdin);  freopen("out.txt", "w", stdout); }
//void FILEIO() {  freopen("namenum.in", "r", stdin);  freopen("namenum.out", "w", stdout); }

int n, a, b, c, d;
string s;

bool check(int a, int b) { // di duoc tu a den b, khong tinh co nguoi o giua
	FOR(i, a, b) {
		char c1 = s[i], c2 = s[i+1];
		if(c1=='#' and c2=='#') return false;
	}
	return true;
}

void solve() {
	cin >> n >> a >> b >> c >> d; a--, b--, c--, d--;
	cin >> s;
	if(!check(a, c) or !check(b, d)) {cout << "No" << endl; return;}
	if(c<d) {cout << "Yes" << endl; return;}
	// now d < c, can go from a to c, from b to d
	FOR(i, b, d) {
		if(s[i]=='.' and s[i+1]=='.' and s[i+2]=='.') {
			cout << "Yes" << endl; return;
		}
	}
	cout << "No" << endl;
}

//#undef int
signed main() {
//#define int long long
  ios_base::sync_with_stdio(false); cin.tie();
//  FILEIO();
//auto beginProgram = chrono::steady_clock::now();

// int t; cin >> t;
//  while (t--)
  {
		solve();
  }

  //auto endProgram = chrono::steady_clock::now();

  return 0;
}