Karen has just arrived at school, and she has a math test today! [Image] 
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 10^9 + 7.
-----Input-----
The first line of input contains a single integer n (1 ≤ n ≤ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is a_{i} (1 ≤ a_{i} ≤ 10^9), the i-th number on the first row.
-----Output-----
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 10^9 + 7.
-----Examples-----
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
-----Note-----
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows: [Image] 
The non-negative remainder after dividing the final number by 10^9 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows: [Image] 
The non-negative remainder after dividing the final number by 10^9 + 7 is 10^9 + 6, so this is the correct output.
#!/usr/bin/env pypy3

import math

def make_nCr_mod(max_n=2*10**5 + 100, mod=10**9 + 7):
    fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)
    fact[0] = 1
    for i in range(max_n):
        fact[i + 1] = fact[i] * (i + 1) % mod

    inv_fact[-1] = pow(fact[-1], mod - 2, mod)
    for i in reversed(range(max_n)):
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod

    def nCr_mod(n, r):
        res = 1
        while n or r:
            a, b = n % mod, r % mod
            if a < b:
                return 0
            res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod
            n //= mod
            r //= mod
        return res

    return nCr_mod

nCr_mod = make_nCr_mod()

MODULUS = 10**9+7

input()
A = input().split(' ')
A = list(map(int, A))

if len(A) == 1:
    print(A[0])
    return

if len(A) % 2 == 1:
    new_A = []
    next_plus = True
    for i in range(len(A) - 1):
        if next_plus:
            new_A += [A[i] + A[i+1]]
        else:
            new_A += [A[i] - A[i+1]]
        next_plus = not next_plus
    A = new_A

if len(A) % 4 == 2:
    new_A = []
    for i in range(len(A) // 2):
        new_A += [A[2*i] + A[2*i+1]]
    A = new_A
else:
    new_A = []
    for i in range(len(A) // 2):
        new_A += [A[2*i] - A[2*i+1]]
    A = new_A

# binomial sum

N = len(A)-1

ret = 0

for i in range(N+1):
    ret += A[i]*nCr_mod(N, i)
    ret = ret % MODULUS

print(ret)
#!/usr/bin/env pypy3

import math

def make_nCr_mod(max_n=2*10**5 + 100, mod=10**9 + 7):
    fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)
    fact[0] = 1
    for i in range(max_n):
        fact[i + 1] = fact[i] * (i + 1) % mod

    inv_fact[-1] = pow(fact[-1], mod - 2, mod)
    for i in reversed(range(max_n)):
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod

    def nCr_mod(n, r):
        res = 1
        while n or r:
            a, b = n % mod, r % mod
            if a < b:
                return 0
            res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod
            n //= mod
            r //= mod
        return res

    return nCr_mod

nCr_mod = make_nCr_mod()

MODULUS = 10**9+7

input()
A = input().split(' ')
A = list(map(int, A))

if len(A) == 1:
    print(A[0])
    return

if len(A) % 2 == 1:
    new_A = []
    next_plus = True
    for i in range(len(A) - 1):
        if next_plus:
            new_A += [A[i] + A[i+1]]
        else:
            new_A += [A[i] - A[i+1]]
        next_plus = not next_plus
    A = new_A

if len(A) % 4 == 2:
    new_A = []
    for i in range(len(A) // 2):
        new_A += [A[2*i] + A[2*i+1]]
    A = new_A
else:
    new_A = []
    for i in range(len(A) // 2):
        new_A += [A[2*i] - A[2*i+1]]
    A = new_A

# binomial sum

N = len(A)-1

ret = 0

for i in range(N+1):
    ret += A[i]*nCr_mod(N, i)
    ret = ret % MODULUS

print(ret)