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;

#define dump(x)  cout << #x << " = " << (x) << endl;

typedef long long ll;
typedef pair<int, int> ii;
const int mod = 1e9+7;
const ll INF = 1e18;
const int inf = 1e9;

void solve()
{
  int n,a,b,c,d;
  cin>>n>>a>>b>>c>>d;
  string s;
  cin>>s;
  vector<ii> f;
  int p=0;
  f.push_back(make_pair(s[0],1));
  for (int i=1; i<n; i++) {
    if (s[i]==f[p].first) f[p].second++;
    else {
      f.push_back(make_pair(s[i],1));
      p++;
    }
  }
  int x = (a<b)^(c<d);
  int mxiwa = 0;
  int mxflt = 0;
  for (auto& pr: f) {
    if (pr.first=='#') mxiwa = max(mxiwa, pr.second);
    else mxflt = max(mxflt, pr.second);
  }

  if (mxiwa>=2) {
    cout << "No" << endl;
    return;
  }

  if (!x) {
    cout << "Yes" << endl;
    return;
  }

  if (mxflt>=3) {
    cout << "Yes" << endl;
    return;
  }

  cout << "No" << endl;
}

int main()
{
  solve();
  return 0;
}