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>
#define int long long
#define rep(i, n) REP(i, 0, n)
#define REP(i, l, r) for (long long i = l; i < r; i++)
using namespace std;
typedef pair<int, int> P;

struct edge {
    int to, cost;
};
class dijkstra {
    int V;
    int d[100002];
    priority_queue<P, vector<P>, greater<P>> que;
    void update(int s, vector<edge> G[]) {
        fill(d, d + V + 1, 100000000000ll);
        d[s] = 0;
        que.push(make_pair(0, s));
        while (!que.empty()) {
            P p = que.top();
            que.pop();
            int v = p.second;
            if (d[v] < p.first) continue;
            rep(i, G[v].size()) {
                edge e = G[v][i];
                if (d[e.to] > d[v] + e.cost) {
                    d[e.to] = d[v] + e.cost;
                    que.push(make_pair(d[e.to], e.to));
                }
            }
        }
    }

public:
    dijkstra(int v, int s, vector<edge> G[]) {
        V = v;
        update(s, G);
    }
    int find(int e) {
        return d[e];
    }
};
int gcd(int a, int b) {
    if (a % b == 0)
        return b;
    else
        return gcd(b, a % b);
}
int lcm(int a, int b) {
    return a * b / gcd(a, b);
}
bool prime(int a) {
    if (a == 1) return false;
    for (int i = 2; i * i <= a; i++) {
        if (a % i == 0) return false;
    }
    return true;
}
class Union_Find {
private:
    vector<int> t, s;

public:
    Union_Find(int max_length) {
        rep(i, max_length + 1) {
            t.push_back(i);
            s.push_back(1);
        }
    }
    void Union(int x, int y) {
        if (same(x, y)) return;
        int tx = Find(x), ty = Find(y);
        if (s[tx] < s[ty]) {
            s[ty] += s[tx];
            t[tx] = ty;
        }
        else if (s[tx] > s[ty]) {
            s[tx] += s[ty];
            t[ty] = tx;
        }
        else if (tx > ty) {
            t[tx] = ty;
            s[ty] += s[tx];
        }
        else {
            t[ty] = tx;
            s[tx] += s[ty];
        }
    }
    int Find(int n) {
        if (t[n] == n)
            return n;
        else
            return t[n] = Find(t[n]);
    }
    bool same(int x, int y) {
        return Find(x) == Find(y);
    }
    int get_Size(int a) {
        return s[a];
    }
};
signed main() {
    int n,a,b,c,d;
    string s;
    cin>>n>>a>>b>>c>>d>>s;
    bool bo=false,boo=false;
    REP(i,2,s.size()){
        if(((a<i&&i<c)||(b<i&&i<d))&&s[i]=='#'&&s[i-1]=='#')boo=true;
        if(d>=i&&a<i&&b<i&&s[i]=='.'&&s[i-1]=='.'&&s[i-2]=='.')bo=true;
    }
    if((bo==false&&d<c)||boo)cout<<"No"<<endl;
    else cout<<"Yes"<<endl;
}