human
Write me a Python code to print the answer in a single line given the following:
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].
Using python3 can you solve the prior task?
assistant
#Here is Python code to print the answer in a single line.

import os
import sys
from io import BytesIO, IOBase

_str = str
str = lambda x=b"": x if type(x) is bytes else _str(x).encode()

BUFSIZE = 8192
class FenwickTree:
    def __init__(self, x):
        """transform list into BIT"""
        self.bit = x
        for i in range(len(x)):
            j = i | (i + 1)
            if j < len(x):
                x[j] += x[i]

    def update(self, idx, x):
        """updates bit[idx] += x"""
        while idx < len(self.bit):
            self.bit[idx] += x
            idx |= idx + 1

    def query(self, end):
        """calc sum(bit[:end])"""
        x = 0
        while end:
            x += self.bit[end - 1]
            end &= end - 1
        return x

    def findkth(self, k):
        """Find largest idx such that sum(bit[:idx]) <= k"""
        idx = -1
        for d in reversed(range(len(self.bit).bit_length())):
            right_idx = idx + (1 << d)
            if right_idx < len(self.bit) and k >= self.bit[right_idx]:
                idx = right_idx
                k -= self.bit[idx]
        return idx + 1

def main():
    mod=10**9+7
    l=[0]*(2*10**5+1)
    ans=1
    for i in range(1,2*10**5+1):
        ans=(ans*i)%mod
        if(i%2==0):
            l[i]=(ans*pow(2,mod-2,mod))%mod
    for _ in range(int(input())):
        n=int(input())
        print(l[2*n])
       



main()
class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd,...