#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;

int N, K, D[2] = { -1, +1 };
string S;

int solve(vi a, int d, int p) {
    int to = 0;
    int res = 0;
    int cnt = 0;

    for (int i = p; i >= 0 && i < K; i+=d) {
        if (a[i]) {
            cnt++;
            to = i;
        }
    }


    for (int i = p; i != to + d; i += d) {
        if (a[i]) {
            if (i == to) {
                a[i] ^= 1;
                res += 3;
            } else {
                a[i] ^= 1;
                a[i + d] ^= 1;
                if (a[i + d])cnt++;
                else cnt--;
                i -= d;
                res += 2;
            }
            cnt--;
        } else {
            if (i != to) {
                a[i + d] ^= 1;
                if (a[i + d])cnt++;
                else cnt--;
                res++;
            }
        }
        if (cnt == 0)break;
    }

    return res;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> S;

    K = sz(S);
    N--;

    int l = -1, r = -1;
    rep(i, K) if (S[i] == '#')r = i;
    for (int i = K - 1; i >= 0; --i)if (S[i] == '#')l = i;

    int P[] = { l,r };

    int ans = INT_MAX;

    rep(x, 2) {
        int p = P[x];

        vi v(K);
        rep(i, K)v[i] = S[i] == '#' ? 1 : 0;

        if (N <= l) {
            smin(ans, solve(v, 1, N));
        } else if (N >= r) {
            smin(ans, solve(v, -1, N));
        } else {

            int d = (p - N) / abs(p - N);
            for (int i = N + d; i != p + d; i += d) {
                v[i] ^= 1;
            }
            smin(ans, solve(v, -D[x], p) + abs(N - p));
        }
    }

    cout << ans << endl;
}