Problem:
Score : 800 points

Problem Statement

Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the 0 key, the 1 key and the backspace key.

To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

The 0 key: a letter 0 will be inserted to the right of the string.

The 1 key: a letter 1 will be inserted to the right of the string.

The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.

Sig has launched the editor, and pressed these keys N times in total. As a result, the editor displays a string s. Find the number of such ways to press the keys, modulo 10^9 + 7.

Constraints

1 ≦ N ≦ 5000

1 ≦ |s| ≦ N

s consists of the letters 0 and 1.

Partial Score

400 points will be awarded for passing the test set satisfying 1 ≦ N ≦ 300.

Input

The input is given from Standard Input in the following format:

N
s

Output

Print the number of the ways to press the keys N times in total such that the editor displays the string s in the end, modulo 10^9+7.

Sample Input 1

3
0

Sample Output 1

5

We will denote the backspace key by B. The following 5 ways to press the keys will cause the editor to display the string 0 in the end: 00B, 01B, 0B0, 1B0, BB0. In the last way, nothing will happen when the backspace key is pressed.

Sample Input 2

300
1100100

Sample Output 2

519054663

Sample Input 3

5000
01000001011101000100001101101111011001000110010101110010000

Sample Output 3

500886057

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;

#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 = (int) 1e9 + 7;

    /// <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

public class Program
{
    public void Solve()
    {
        var sc = new Scanner();
        int N = sc.NextInt();
        string S = sc.Next();

        var dp = new ModInt[N + 1, N + 1];
        dp[0, 0] = 1;

        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j <= i; j++)
            {
                dp[i + 1, j + 1] += dp[i, j] * 2;
                dp[i + 1, Math.Max(0, j - 1)] += dp[i, j];
            }
        }

        ModInt ans = dp[N, S.Length] * ModInt.Inverse(ModInt.Pow(2, S.Length));
        Console.WriteLine(ans);
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}