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>
using namespace std;
using ll = long long;
using Edge = pair<int, int>;
using Graph = vector<vector<Edge>>;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define RSORT(v) sort((v).rbegin(), (v).rend())
const ll MOD = 1000000007;
const ll nmax = 8;
const ll INF = 1e9;
bool graph[nmax][nmax];
vector<vector<ll>> dist = vector<vector<ll>>(nmax, vector<ll>(nmax, INF));
void warshall_floyd(ll n)
{
    for (size_t i = 0; i < n; i++)
    {
        for (size_t j = 0; j < n; j++)
        {
            for (size_t k = 0; k < n; k++)
            {
                dist[j][k] = min(dist[j][k], dist[j][i] + dist[i][k]);
            }
        }
    }
}

class UnionFind
{
public:
    vector<ll> Parent;

    UnionFind(ll N)
    {
        Parent = vector<ll>(N, -1);
    }
    ll find(ll A)
    {
        if (Parent[A] < 0)
            return A;
        return Parent[A] = find(Parent[A]);
    }

    ll size(ll A)
    {
        return -Parent[find(A)];
    }

    bool Union(ll A, ll B)
    {
        A = find(A);
        B = find(B);
        if (A == B)
        {
            return false;
        }
        if (size(A) < size(B))
            swap(A, B);

        Parent[A] += Parent[B];
        Parent[B] = A;

        return true;
    }
};

ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    ll g = gcd(a, b);
    return a / g * b;
}

ll mulMod(ll a, ll b)
{
    return (((a % MOD) * (b % MOD)) % MOD);
}

ll powMod(ll a, ll p)
{
    if (p == 0)
    {
        return 1;
    }
    else if (p % 2 == 0)
    {
        ll half = powMod(a, p / 2);
        return mulMod(half, half);
    }
    else
    {
        return mulMod(powMod(a, p - 1), a);
    }
}

ll ceil(ll a, ll b)
{
    return (a + b - 1) / b;
}

const string YES = "Yes";
const string NO = "No";

void solve(long long N, long long A, long long B, long long C, long long D, std::string S)
{

    for (int i = min(A - 1, B - 1); i < max(C - 1, D - 1) - 1; i++)
    {
        if (S[i] == '#' && S[i + 1] == '#')
        {
            cout << NO << endl;
            return;
        }
    }

    if (C < D)
    {
        cout << YES << endl;
        return;
    }
    else
    {
        for (int i = max(A - 1, B - 1); i < min(C - 1, D - 1) - 2; i++)
        {
            if (S[i] == '.' && S[i + 1] == '.' && S[i + 2] == '.')
            {
                cout << YES << endl;
                return;
            }
        }
        cout << NO << endl;
    }
}

int main()
{
    long long N;
    scanf("%lld", &N);
    long long A;
    scanf("%lld", &A);
    long long B;
    scanf("%lld", &B);
    long long C;
    scanf("%lld", &C);
    long long D;
    scanf("%lld", &D);
    std::string S;
    std::cin >> S;
    solve(N, A, B, C, D, S);
    return 0;
}