Problem:
Score : 800 points

Problem Statement

Given are integers N and X. For each integer k between 0 and X (inclusive), find the answer to the following question, then compute the sum of all those answers, modulo 998244353.

Let us repeat the following operation on the integer k. Operation: if the integer is currently odd, subtract 1 from it and divide it by 2; otherwise, divide it by 2 and add 2^{N-1} to it. How many operations need to be performed until k returns to its original value? (The answer is considered to be 0 if k never returns to its original value.)

Constraints

1 \leq N \leq 2\times 10^5

0 \leq X < 2^N

X is given in binary and has exactly N digits. (In case X has less than N digits, it is given with leading zeroes.)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print the sum of the answers to the questions for the integers between 0 and X (inclusive), modulo 998244353.

Sample Input 1

3
111

Sample Output 1

40

For example, for k=3, the operation changes k as follows: 1,0,4,6,7,3. Therefore the answer for k=3 is 6.

Sample Input 2

6
110101

Sample Output 2

616

Sample Input 3

30
001110011011011101010111011100

Sample Output 3

549320998

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T> using treap = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define int long long
#define REP(i, n) for (long long i = 0, max_i = (n); i < max_i; i++)
#define REPI(i, a, b) for (long long i = (a), max_i = (b); i < max_i; i++)
#define ALL(obj) begin(obj), end(obj)
#define RALL(obj) rbegin(obj), rend(obj)
#define fi first
#define se second
using ii = pair<int, int>;
vector<ii> dirs = {
    {1, 0}, {0, 1}, {-1, 0}, {0, -1},  // 4方向
    {1, 1}, {-1, 1}, {-1, -1}, {1, -1},  // 斜め
    {0, 0},  // 自身
};
template <class T> inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } return false; }
template <class T> inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; } return false; }
template <class T, class S> vector<T> make_vec(size_t n, S x) { return vector<T>(n, x); }
template <class T, class... Ts> auto make_vec(size_t n, Ts... ts) { return vector<decltype(make_vec<T>(ts...))>(n, make_vec<T>(ts...)); }

// debug
template <class T> ostream& operator<<(ostream& s, vector<T>& d) { REP (i, d.size()) s << d[i] << (i == d.size() - 1 ? "" : " "); return s; }
template <class T> ostream& operator<<(ostream& s, vector<vector<T>>& d) { REP (i, d.size()) s << d[i] << (i == d.size() - 1 ? "" : "\n"); return s; }
template <class T, class S> ostream& operator<<(ostream& s, pair<T, S>& p) { s << "{" << p.first << ", " << p.second << "}"; return s; }
template <class T> ostream& operator<<(ostream& s, set<T> m) { for (auto it = m.begin(); it != m.end(); it++) { s << *it << (next(it) == m.end() ? "" : "\n"); } return s; }
template <class T> ostream& operator<<(ostream& s, multiset<T> m) { for (auto it = m.begin(); it != m.end(); it++) { s << *it << (next(it) == m.end() ? "" : "\n"); } return s; }
template <class T, class S> ostream& operator<<(ostream& s, map<T, S> m) { for (auto it = m.begin(); it != m.end(); it++) { s << *it << (next(it) == m.end() ? "" : "\n"); } return s; }
template <class T, class S> ostream& operator<<(ostream& s, unordered_map<T, S> m) { for (auto it = m.begin(); it != m.end(); it++) { s << *it << (next(it) == m.end() ? "" : "\n"); } return s; }
#ifdef _MY_DEBUG
    #define dump(...) cerr << "/* " << #__VA_ARGS__ << " :[" << __LINE__ << ":" << __FUNCTION__ << "]" << endl, dump_func(__VA_ARGS__), cerr << "*/\n\n";
#else
    #define dump(...)
    #define endl "\n"
#endif
void dump_func() { cerr << endl; }
template <class Head, class... Tail> void dump_func(Head&& h, Tail&&... t) { cerr << h << (sizeof...(Tail) == 0 ? "" : ", "), dump_func(forward<Tail>(t)...); }

struct Fast { Fast() { cin.tie(nullptr); ios::sync_with_stdio(false); } } fast;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
constexpr int MOD = 998244353;
// *************** TEMPLATE END *************** 

template <class T>
T pow(T x, int n, const T UNION = 1) {
    T ret = UNION;
    while (n) {
        if (n & 1) ret *= x;
        x *= x;
        n >>= 1;
    }
    return ret;
}

template <int MD>
struct ModInt {
    int x;
    static unordered_map<int, int> to_inv;
    ModInt() : x(0) {}
    ModInt(int x_) {
        if ((x = x_ % MD + MD) >= MD) x -= MD;
    }

    ModInt& operator+=(ModInt that) {
        if ((x += that.x) >= MD) x -= MD;
        return *this;
    }
    ModInt& operator*=(ModInt that) {
        x = (unsigned long long)x * that.x % MD;
        return *this;
    }
    ModInt& operator-=(ModInt that) {
        if ((x -= that.x) < 0) x += MD;
        return *this;
    }
    ModInt& operator/=(ModInt that) {
        x = (unsigned long long)x * that.inv().x % MD;
        return *this;
    }

    ModInt operator-() const { return -x < 0 ? MD - x : -x; }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt inv() const { return to_inv.count(this->x) ? to_inv[this->x] : (to_inv[this->x] = pow(*this, MD - 2).x); }
    friend ostream& operator<<(ostream& s, ModInt<MD> a) {
        s << a.x;
        return s;
    }
    friend istream& operator>>(istream& s, ModInt<MD>& a) {
        s >> a.x;
        return s;
    }
};
template <int MD>
unordered_map<int, int> ModInt<MD>::to_inv;
using mint = ModInt<MOD>;

vector<mint> fact, fact_inv;
void init_factorial(int n) {
    fact = vector<mint>(n + 1, 1);
    fact_inv = vector<mint>(n + 1);
    for (int i = 0; i < n; i++) fact[i + 1] = fact[i] * (i + 1);
    fact_inv[n] = mint(1) / fact[n];
    for (int i = n - 1; i >= 0; i--) fact_inv[i] = fact_inv[i + 1] * (i + 1);
    // for (int i = 0; i < n + 1; i++) assert(fact[i] * fact_inv[i] == 1);
}
mint comb(int n, int r) { return fact[n] * fact_inv[r] * fact_inv[n - r]; }

map<int, int> prime_factorization(int n) {
    map<int, int> mp;
    for (int x = 2; x * x <= n; x++) {
        while (n % x == 0) {
            mp[x]++;
            n /= x;
        }
    }
    if (n != 1) mp[n]++;
    return mp;
}
signed main() {
    int n; cin >> n;
    string s; cin >> s;
    vector<int> ps;
    for (int x = 1; x * x <= n; x++) {
        if (n % x) continue;
        int y = n / x;
        if (x % 2) ps.push_back(y);
        if (x != y && y % 2) ps.push_back(x);
    }
    sort(ALL(ps));

    auto check = [&](int y) {
        vector<string> st = {s.substr(0, y), ""};
        for (char c : st[0]) st[1].push_back(c == '0' ? '1' : '0');
        for (int i = y; i < n; i += y) {
            swap(st[0], st[1]);
            string tmp = s.substr(i, y);
            if (tmp > st[0]) return true;
            if (tmp < st[0]) return false;
        }
        return true;
    };
    mint ans = 0;
    int m = ps.size();
    vector<mint> num(m);
    REP (i, m) {
        REP (k, ps[i]) {
            num[i] *= 2;
            num[i] += s[k] == '1';
        }
        num[i] += check(ps[i]);
        REP (j, i) {
            if (ps[i] % ps[j] == 0) num[i] -= num[j];
        }
        ans += num[i] * (2 * ps[i]);
    }
    cout << ans << endl;
}