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 <ios>
#include <iomanip>
#include <iostream>
#include <functional>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <climits>
 
using namespace std;
typedef long long int ll;
typedef long double ld;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define INF (1e9)

#define PI (acos(-1))
#define print(x) cout << x << endl
 
ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
typedef pair <ll,ll> P;

int main()
{
    ll N, A, B, C, D;
    string S;
    cin >> N >> A >> B >> C >> D >> S;
    vector<bool> vec(N, true);
    rep(i, N) {
        if (S[i] == '#') {
            vec[i] = false;
        }
    }
    if (C > D) {
        bool flag1 = false;
        for(ll i = A-1; i < C-1; i++) {
            if (!vec[i] && !vec[i+1]) {
                flag1 = true;
            }
        }
        for(ll i = B-1; i < D-1; i++) {
            if (!vec[i] && !vec[i+1]) {
                flag1 = true;
            }
        }
        bool flag2 = false;
        for(ll i = B-1; i < D-1; i++) {
            if (vec[i] && vec[i+1] && vec[i+2]) {
                flag2 = true;
            }
        }
        if (!flag1 && flag2) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
    } else {
        bool flag = false;
        for(ll i = A-1; i < C-1; i++) {
            if (!vec[i] && !vec[i+1]) {
                flag = true;
            }
        }
        for(ll i = B-1; i < D-1; i++) {
            if (!vec[i] && !vec[i+1]) {
                flag = true;
            }
            }
        if (flag) {
            cout << "No" << endl;
        } else {
            cout << "Yes" << endl;
        }
    }
    return 0;
}