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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Diagnostics;
using Debug = System.Diagnostics.Trace;
using SB = System.Text.StringBuilder;
using static System.Math;
using Number = System.Int64;
using System.Numerics;
using static System.Numerics.BigInteger;
namespace Program {
	public class Solver {
		Random rnd = new Random(0);
		public void Solve() {
			var n = ri;
			var s = rs;

			var cnt = new ModInt[2 * n + 1];
			for (int d = 2; d <= 2 * n; d += 2)
				if (2 * n % d == 0 && n % d == d / 2) {
					ModInt just = 1;
					ModInt lower = 0;
					var D = d / 2;
					for (int i = 0; i < n; i++) {
						if (i < D) {
							lower *= 2;
							if (s[i] == '1') lower += just;
						}
						else {
							var p = s[i % D] - '0';
							if ((i / D) % 2 == 1) p ^= 1;
							var v = s[i] - '0';
							if (p > v) just = 0;
							else if (p < v) break;
						}
					}
					cnt[d] += just + lower;
					Debug.WriteLine($"{d} {cnt[d]}");
					for (int j = 1; j < d; j++)
						if (d % j == 0) cnt[d] -= cnt[j];
				}
			Debug.WriteLine(cnt.AsJoinedString());
			ModInt ans = 0;
			for (int i = 0; i <= 2 * n; i++)
				ans += i * cnt[i];
			Console.WriteLine(ans);
		}
		const long INF = 1L << 60;
		static int[] dx = { -1, 0, 1, 0 };
		static int[] dy = { 0, 1, 0, -1 };
		int ri { get { return sc.Integer(); } }
		long rl { get { return sc.Long(); } }
		double rd { get { return sc.Double(); } }
		string rs { get { return sc.Scan(); } }
		public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());
		static T[] Enumerate<T>(int n, Func<int, T> f) {
			var a = new T[n]; for (int i = 0; i < a.Length; ++i) a[i] = f(i); return a;
		}
		static T[][] Enumerate<T>(int n, int m, Func<int, int, T> f) {
			return Enumerate(n, x => Enumerate(m, y => f(x, y)));
		}
		static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
	}
}

#region main
static class Ex {
	static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
	static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
		return string.Join(st, ie);
	}
	static public void Main() {
		Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = false });
		var solver = new Program.Solver();
		//* 
		var t = new System.Threading.Thread(solver.Solve, 200000000);
		t.Start();
		t.Join();
		//*/
		// solver.Solve();
		Console.Out.Flush();
	}
}
#endregion
#region Ex
namespace Program.IO {
	using System.IO;
	using System.Text;
	using System.Globalization;

	public class Printer : StreamWriter {
		public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
		public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
	}

	public class StreamScanner {
		public StreamScanner(Stream stream) { str = stream; }

		public readonly Stream str;
		private readonly byte[] buf = new byte[1024];
		private int len, ptr;
		public bool isEof = false;
		public bool IsEndOfStream { get { return isEof; } }

		private byte read() {
			if (isEof) return 0;
			if (ptr >= len) {
				ptr = 0;
				if ((len = str.Read(buf, 0, 1024)) <= 0) {
					isEof = true;
					return 0;
				}
			}
			return buf[ptr++];
		}

		public char Char() {
			byte b = 0;
			do b = read(); while ((b < 33 || 126 < b) && !isEof);
			return (char)b;
		}
		public string Scan() {
			var sb = new StringBuilder();
			for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
			return sb.ToString();
		}
		public string ScanLine() {
			var sb = new StringBuilder();
			for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
			return sb.ToString();
		}
		public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
		public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
		public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
	}
}

#endregion
#region Disjoint Set
/// <summary>
/// いくつかの要素を素集合に分割して保持するコレクションです．
/// </summary>
public class DisjointSet {
	int[] par;
	byte[] rank;
	/// <summary>
	/// 0 から <paramref name="N"/>-1 までの番号がついた <paramref name="N"/> 個の要素からなる集合を作成します．
	/// </summary>
	/// <param name="N">要素数</param>
	/// <remarks>このコンストラクタは O(N) で実行されます．</remarks>
	public DisjointSet(int N) {
		par = new int[N];
		rank = new byte[N];
		for (int i = 0; i < N; i++)
			par[i] = -1;
	}
	/// <summary>
	/// 指定した要素が属する集合の代表値を取得します．
	/// </summary>
	/// <param name="id">調べたい要素の 0-indexed での番号</param>
	/// <returns>指定した要素が属する集合の代表値</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public int this[int id] {
		get {
			if ((par[id] < 0)) return id;
			return par[id] = this[par[id]];
		}
	}
	/// <summary>
	/// 指定した 2 つの要素が属する集合同士を 1 つに統合することを試みます．
	/// </summary>
	/// <param name="x">最初の要素の 0-indexed での番号</param>
	/// <param name="y">2 つ目の要素の 0-indexed での番号</param>
	/// <returns>統合に成功したならば true，そうでなければ false．</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public bool Unite(int x, int y) {
		x = this[x]; y = this[y];
		if (x == y) return false;
		if (rank[x] < rank[y]) { var tmp = x; x = y; y = tmp; }
		par[x] += par[y];
		par[y] = x;
		if (rank[x] == rank[y])
			rank[x]++;
		return true;
	}
	/// <summary>
	/// 指定した要素が属する集合のサイズを取得します．
	/// </summary>
	/// <param name="x">指定する要素の 0-indexed での番号</param>
	/// <returns>集合のサイズ</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public int Size(int x) { return -par[this[x]]; }

	/// <summary>
	/// 指定した 2 つの要素が属する集合が同じかどうかを判定する．計算量 O(α(N))．
	/// </summary>
	/// <param name="x">最初の要素の 0-indexed での番号</param>
	/// <param name="y">2 つ目の要素の 0-indexed での番号</param>
	/// <returns>同じならば true，そうでなければ false．</returns>
	/// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
	public bool Same(int x, int y) { return this[x] == this[y]; }

}
#endregion
#region ModInt
/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt {
	/// <summary>
	/// 剰余を取る値．
	/// </summary>
	public const long Mod = 998244353;

	/// <summary>
	/// 実際の数値．
	/// </summary>
	public long num;
	/// <summary>
	/// 値が <paramref name="n"/> であるようなインスタンスを構築します．
	/// </summary>
	/// <param name="n">インスタンスが持つ値</param>
	/// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
	public ModInt(long n) { num = n; }
	/// <summary>
	/// このインスタンスの数値を文字列に変換します．
	/// </summary>
	/// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
	public override string ToString() { return num.ToString(); }
	public static ModInt operator +(ModInt l, ModInt r) { l.num += r.num; if (l.num >= Mod) l.num -= Mod; return l; }
	public static ModInt operator -(ModInt l, ModInt r) { l.num -= r.num; if (l.num < 0) l.num += Mod; return l; }
	public static ModInt operator *(ModInt l, ModInt r) { return new ModInt(l.num * r.num % Mod); }
	public static implicit operator ModInt(long n) { n %= Mod; if (n < 0) n += Mod; return new ModInt(n); }

	/// <summary>
	/// 与えられた 2 つの数値からべき剰余を計算します．
	/// </summary>
	/// <param name="v">べき乗の底</param>
	/// <param name="k">べき指数</param>
	/// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
	public static ModInt Pow(ModInt v, long k) { return Pow(v.num, k); }

	/// <summary>
	/// 与えられた 2 つの数値からべき剰余を計算します．
	/// </summary>
	/// <param name="v">べき乗の底</param>
	/// <param name="k">べき指数</param>
	/// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
	public static ModInt Pow(long v, long k) {
		long ret = 1;
		for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
			if ((k & 1) == 1) ret = ret * v % Mod;
		return new ModInt(ret);
	}
	/// <summary>
	/// 与えられた数の逆元を計算します．
	/// </summary>
	/// <param name="v">逆元を取る対象となる数</param>
	/// <returns>逆元となるような値</returns>
	/// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
	public static ModInt Inverse(ModInt v) { return Pow(v, Mod - 2); }
}
#endregion