Problem: 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`.

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.

Examples

Input

3
0


Output

5


Input

300
1100100


Output

519054663


Input

5000
01000001011101000100001101101111011001000110010101110010000


Output

500886057

Solution (in PYTHON3):

Solution:

"""
Writer: SPD_9X2
https://atcoder.jp/contests/arc059/tasks/arc059_d

dp[i][j] = i回キーボードを押した時に、j文字の通り数
とすれば、O(N**2)で解ける。

0,1は等確率で現れることを考えれば、文字数を減らした場合のdp推移も簡単。
2(mod10**9+7)の逆元が必要かな→いらんかった

(最後に|s|文字になっている押し方/2**|s|)通りが求める答え
"""

def inverse(a,mod): #aのmodを法にした逆元を返す
    return pow(a,mod-2,mod)


N = int(input())

s = input()

dp = [0] * (N+1)
dp[0] = 1

mod = 10**9+7

for loop in range(N):

    ndp = [0] * (N+1)

    for i in range(N+1):

        dp[i] %= mod
        if i != N:
            ndp[i+1] += dp[i] * 2
            ndp[i+1] %= mod

        if i != 0:
            ndp[i-1] += dp[i]
            ndp[i-1] %= mod
        else:
            ndp[i] += dp[i]
            ndp[i] %= mod

    dp = ndp

    #print (dp)
inv = inverse(pow(2,len(s),mod) , mod)
print (dp[len(s)] * inv % mod)