CQXYM is counting permutations length of 2n.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

A permutation p(length of 2n) will be counted only if the number of i satisfying p_i<p_{i+1} is no less than n. For example:

  * Permutation [1, 2, 3, 4] will count, because the number of such i that p_i<p_{i+1} equals 3 (i = 1, i = 2, i = 3).
  * Permutation [3, 2, 1, 4] won't count, because the number of such i that p_i<p_{i+1} equals 1 (i = 3). 



CQXYM wants you to help him to count the number of such permutations modulo 1000000007 (10^9+7).

In addition, [modulo operation](https://en.wikipedia.org/wiki/Modulo_operation) is to get the remainder. For example:

  * 7 mod 3=1, because 7 = 3 ⋅ 2 + 1, 
  * 15 mod 4=3, because 15 = 4 ⋅ 3 + 3. 

Input

The input consists of multiple test cases. 

The first line contains an integer t (t ≥ 1) — the number of test cases. The description of the test cases follows.

Only one line of each test case contains an integer n(1 ≤ n ≤ 10^5).

It is guaranteed that the sum of n over all test cases does not exceed 10^5

Output

For each test case, print the answer in a single line.

Example

Input


4
1
2
9
91234


Output


1
12
830455698
890287984

Note

n=1, there is only one permutation that satisfies the condition: [1,2].

In permutation [1,2], p_1<p_2, and there is one i=1 satisfy the condition. Since 1 ≥ n, this permutation should be counted. In permutation [2,1], p_1>p_2. Because 0<n, this permutation should not be counted.

n=2, there are 12 permutations: [1,2,3,4],[1,2,4,3],[1,3,2,4],[1,3,4,2],[1,4,2,3],[2,1,3,4],[2,3,1,4],[2,3,4,1],[2,4,1,3],[3,1,2,4],[3,4,1,2],[4,1,2,3].
import math
import sys
import operator as op
from functools import reduce
from collections import OrderedDict


class Input:
    @staticmethod
    def string():
        return str(input())

    @staticmethod
    def int():
        return int(input())

    @staticmethod
    def list(sep=' '):
        return input().split(sep)

    @staticmethod
    def list_int(sep=' '):
        return list(map(int, input().split(sep)))


class Output:
    @staticmethod
    def print(res):
        print(res)

    @staticmethod
    def string(res):
        print(str(res))

    @staticmethod
    def list(res):
        print(' '.join(str(x) for x in res))


class Algorithms:
    @staticmethod
    def isPrime(n):
        return all(n % i != 0 for i in range(2, int(math.sqrt(n) + 1)))

    @staticmethod
    def getPrimeNumbers(n):
        a = [1]
        for num in range(2, n):
            if all(num % i != 0 for i in range(2, int(math.sqrt(num) + 1))):
                a.append(num)
        return a

    # Факторизация - Разложение числа на простые множители
    @staticmethod
    def factorization(n):
        res = {}
        for i in range(2, n * n):
            cnt = 0
            while n % i == 0:
                cnt += 1
                n = n / i
            if cnt > 0:
                res[i] = cnt
        if n > 1:
            res[n] = 1
        return res

    # Наибольший общий делитель (НОД)
    @staticmethod
    def gcd(a, b):
        if b == 0:
            return a
        else:
            a = a % b
            return Algorithms.gcd(b, a)

    # Наименьшее общее кратное (НОК)
    @staticmethod
    def lcm(a, b):
        return abs(a * b) // Algorithms.gcd(a, b)

    # Последовательность Фибоначчи =  0, 1, 1, 2, 3, 5, 8, 13, 21, 34
    @staticmethod
    def fibonacci(n):
        a = [0, 1]
        for i in range(2, n + 1):
            a.append(a[i - 1] - a[i - 2])
        return a[n]

    @staticmethod
    def z_func(s):
        n = len(s)
        z = [0] * n
        l = r = 0
        for i in range(1, n):
            if r >= i:
                z[i] = min(z[i - l], r - i + 1)
            while z[i] + i < n and s[z[i]] == s[z[i] + i]:
                z[i] += 1
            if i + z[i] - 1 > r:
                l = i
                r = i + z[i] - 1
        return z

    @staticmethod
    def faktorial(n):
        factorial = 1
        for i in range(3, n + 1):
            factorial *= i
            factorial = factorial % (math.pow(10, 9) + 7)

        return factorial

def solve():
    n = Input.int()
    p = 2 * n
    print(int(Algorithms.faktorial(p)))


#query_count = 1
query_count = Input.int()
while query_count:
    query_count -= 1
    solve()