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 fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

#define M 998244353

#define ll long long
#define ld long double
#define vi vector<ll>
#define pi pair<ll, ll>
#define vii vector<pi>
#define vvi vector<vi>
#define pb push_back
#define endl "\n"

#define REP(i, s, e) for (ll i = s; i < e; i++)
#define RREP(i, s, e) for (ll i = s; i > e; i--)

#define all(v) v.begin(), v.end()
#define part(v, s, e) v.begin() + s, v.begin() + e
#define print(v) for (auto i: v) cout << i << " ";

int main() {
    fast;
    ll n, a, b, c, d;
    cin >> n >> a >> b >> c >> d;
    a--, b--, c--, d--;
    string s;
    cin >> s;
    if (c < d)
        cout << "Yes\n";
    else {
        ll p = 0;
        REP(i, b-1, d) {
            if (s[i] == '.' && s[i+1] == '.' && s[i+2] == '.') {
                p = 1;
                break;
            }
        }
        if (p)
            cout << "Yes\n";
        else
            cout << "No\n";
    }
    return 0;
}