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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

// AtCoder.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//

// AtCoder.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//

#ifndef ONLINE_JUDGE

#endif
#define _USE_MATH_DEFINES
#include<math.h>

//#include<cmath>

#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
#pragma GCC optimize("Ofast")


//#include "Ants.h"
using namespace std;
typedef long long ll;
#define rad_to_deg(rad) (((rad)/2/M_PI)*360)
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i,n) for(int i=0;i<n;i++)
#define show(s) cout<<s<<endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define LINF (1LL << 50)
#define MOD (1e9+7)
#define rep1(i,n) for(int i =1;i<=n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;--i)

typedef pair < ll, ll> P;
ll gcd(ll a, ll b) {//最大公約数を求める///最大公倍数a*b/gcd(a,b)
	if (b == 0) return a;
	return gcd(b, a % b);
}
ll lcd(ll a, ll b) {//最小公倍数を求める　b=0のときaを返す
	if (b == 0) return a;
	return (a / gcd(a, b)) * b;
}
vector<int> divnum(ll num) {
	int dig;
	vector<int>p;
	while (num) {
		dig = num % 10;
		p.push_back(dig);
		num /= 10;
	}
	return p;
}
int digiter(ll num) {
	int dig;
	vector<int>p;
	while (num) {
		dig = num % 10;
		p.push_back(dig);
		num /= 10;
	}
	return p.size();
}
vector<int> convertstring(string s) {
	vector<int> d;
	ll n = s.size();
	rep(i, n) {
		d.push_back(s[i] - '0');
	}
	return d;
}
map< ll, int > prime_factor(ll n) {
	map< ll, int > ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}

vector<ll> Enumeration(ll n) {
	vector<ll>res;
	for (ll i = 1; i * i <= n; i++) {
		if (n % i != 0)continue;
		res.push_back(i);
		if (i * i != n)res.push_back(n / i);
	}
	
	return res;
}



ll myPow(ll x, ll n, ll m) {//x^n
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}
std::vector<int> Eratosthenes(const int N)//素数列挙
{
	std::vector<bool> is_prime(N + 1);
	for (int i = 0; i <= N; i++)
	{
		is_prime[i] = true;
	}
	std::vector<int> P;
	for (int i = 2; i <= N; i++)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i; j <= N; j += i)
			{
				is_prime[j] = false;
			}
			P.emplace_back(i);
		}
	}
	return P;
}
const ll mod = MOD;

struct mint {

	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x% mod + mod) % mod) {}
	
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
	mint operator+(const mint a) const { return mint(*this) += a; }
	mint operator-(const mint a) const { return mint(*this) -= a; }
	mint operator*(const mint a) const { return mint(*this) *= a; }
	
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const { return pow(mod - 2); }
	mint& operator/=(const mint a) { return *this *= a.inv(); }
	mint operator/(const mint a) const { return mint(*this) /= a; }
	
	//operator long long() const { return x; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct combination {//comb(a,b)でよぶaCb

	vector<mint> fact, ifact;
	combination(ll n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (ll i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (ll i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(ll n, ll k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
}comb(1000005);//comb貯め
mint f(ll n, ll k) {
	if (n < 0) return 0;
	// nPk = nCk * k!
	mint res = comb(n, k);
	res *= comb.fact[k];
	return res;
}




struct SegmentTree {
private:
	int n;
	vector<int> node;

public:
	SegmentTree(vector<int> v) {//SegmentTree seg(vector<int>(N,INF))で呼び出し 
		int sz = v.size();
		n = 1; while (n < sz) n *= 2;
		node.resize(2 * n - 1, INF);
		for (int i = 0; i < sz; i++) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = min(node[2 * i + 1], node[2 * i + 2]);
	}

	void update(int x, int val) {
		x += (n - 1);
		node[x] = val;
		while (x > 0) {
			x = (x - 1) / 2;
			node[x] = min(node[2 * x + 1], node[2 * x + 2]);
		}
	}

	int getmin(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		if (r <= a || b <= l) return INF;
		if (a <= l && r <= b) return node[k];

		int vl = getmin(a, b, 2 * k + 1, l, (l + r) / 2);
		int vr = getmin(a, b, 2 * k + 2, (l + r) / 2, r);
		return min(vl, vr);
	}
};
// 1-indexedなので注意。
struct BIT {
private:
	vector<int> bit;
	int N;
public:
	BIT(int size) {
		N = size;
		bit.resize(N + 1);
	}

	// 一点更新です
	void add(int a, int w) {//a番目をwに更新
		for (int x = a; x <= N; x += x & -x) bit[x] += w;
	}

	// 1~Nまでの和を求める。
	int sum(int a) {
		int ret = 0;
		for (int x = a; x > 0; x -= x & -x) ret += bit[x];
		return ret;
	}
};
/****Union-Ford-Tree***/
/*int par[int(0x7ffffff)];//Parent
int Rank[int(0x7ffffff)];//Deep_of_the_Tree

//int a[100] = { 4,2,3,1,5 }, m[100] ={3,2,2};
//int n = 0;
//const ll mod = 1000000007;


//n要素で初期化
void init(int n) {
	for (int i = 0; i < n; i++) {
		par[i] = i;
		Rank[i] = 0;
	}
}
//木の根探し
int find(int x) {
	if (par[x] == x) {
		return x;
	}
	else {
		return par[x] = find(par[x]);
	}
}
//xとyの属する集合を併合
void unite(int x, int y) {
	x = find(x);
	y = find(y);
	if (x == y)return;

	if (Rank[x] < Rank[y]) {
		par[x] = y;
	}
	else {
		par[y] = x;
		if (Rank[x] == Rank[y])Rank[x]++;
	}
}

bool same(int x, int y) {
	return find(x) == find(y);
}*/
//queue古い順,stack新しい順
/************************************************************************************/
int N, A, B, C, D; string S;
int32_t main() {
	cin >> N >> A >> B >> C >> D >> S;
	S += '#'+S+'#';
	auto can_reach = [&](int s, int t) {
		for (int i = s; i < t; i++) {
			if (S[i] == '#' && S[i] == S[i + 1])return false;
		}
		return true;
	};
	/*if (D > C) {
		if (can_reach(A, C) && can_reach(B, D)) {
			cout << "Yes" << endl;
		}
		else cout << "No" << endl;
		return 0;
	}*/
	if (!can_reach(A, C) || !can_reach(B, D)) {
		cout << "No" << endl;
		return 0;
	}
	if (C > D) {

		bool snuke_can_over = false;
		for (int i = B; i <= D; i++) {
			if (S[i - 1] == '.' && S[i] == '.' && S[i + 1] == '.') {
				snuke_can_over = true;
			}
		}
		if (!snuke_can_over) {
			cout << "No" << endl;
			return 0;
		}
	}
	cout << "Yes" << endl;

}