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 <algorithm>
#include <bits/stdc++.h>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;

#define rep(X, S, E) for (int(X) = (S); (X) < (E); ++(X))
#define rrep(X, S, E) for (int(X) = (E)-1; (X) >= (S); --(X))
#define itrep(X, Y) for (auto(X) = (Y).begin(); (X) != (Y).end(); (X)++)
#define all(X) (X).begin(), (X).end()
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
#define print(x) cout << x << endl

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ll INF = 9 * 1e18;
const ll MOD = 1e9 + 7;

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

string S;

bool isOK(ll from, ll to) {
  rep(i, from + 1, to - 1) {
    if (S[i] == '#' && S[i + 1] == '#') {
      return false;
    }
  }
  return true;
}

bool isOK2(ll from, ll to, ll middle) {
  if (S[middle - 1] == '.' && S[middle + 1] == '.') {
    return true;
  }

  rep(i, middle + 1, to - 1) {
    if (S[i] == '.' && S[i + 1] == '.') {
      return true;
    }
  }

  return false;
}

void solve(long long N, long long A, long long B, long long C, long long D) {
  A--;
  B--;
  C--;
  D--;

  if (D < C) {
    if (isOK(B, D) && isOK(A, C) && isOK2(A, C, B)) {
      print(YES);
      return;
    }
  } else {
    if (isOK(B, D) && isOK(A, C)) {
      print(YES);
      return;
    }
  }
  print(NO);
}

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::cin >> S;
  solve(N, A, B, C, D);
  return 0;
}