Dr. Evil is interested in math and functions, so he gave Mahmoud and Ehab array a of length n and array b of length m. He introduced a function f(j) which is defined for integers j, which satisfy 0 ≤ j ≤ m - n. Suppose, c_{i} = a_{i} - b_{i} + j. Then f(j) = |c_1 - c_2 + c_3 - c_4... c_{n}|. More formally, $f(j) =|\sum_{i = 1}^{n}(- 1)^{i - 1} *(a_{i} - b_{i + j})|$. 

Dr. Evil wants Mahmoud and Ehab to calculate the minimum value of this function over all valid j. They found it a bit easy, so Dr. Evil made their task harder. He will give them q update queries. During each update they should add an integer x_{i} to all elements in a in range [l_{i};r_{i}] i.e. they should add x_{i} to a_{l}_{i}, a_{l}_{i} + 1, ... , a_{r}_{i} and then they should calculate the minimum value of f(j) for all valid j.

Please help Mahmoud and Ehab.


-----Input-----

The first line contains three integers n, m and q (1 ≤ n ≤ m ≤ 10^5, 1 ≤ q ≤ 10^5) — number of elements in a, number of elements in b and number of queries, respectively.

The second line contains n integers a_1, a_2, ..., a_{n}. ( - 10^9 ≤ a_{i} ≤ 10^9) — elements of a.

The third line contains m integers b_1, b_2, ..., b_{m}. ( - 10^9 ≤ b_{i} ≤ 10^9) — elements of b.

Then q lines follow describing the queries. Each of them contains three integers l_{i} r_{i} x_{i} (1 ≤ l_{i} ≤ r_{i} ≤ n,  - 10^9 ≤ x ≤ 10^9) — range to be updated and added value.


-----Output-----

The first line should contain the minimum value of the function f before any update.

Then output q lines, the i-th of them should contain the minimum value of the function f after performing the i-th update .


-----Example-----
Input
5 6 3
1 2 3 4 5
1 2 3 4 5 6
1 1 10
1 1 -9
1 5 -1

Output
0
9
0
0



-----Note-----

For the first example before any updates it's optimal to choose j = 0, f(0) = |(1 - 1) - (2 - 2) + (3 - 3) - (4 - 4) + (5 - 5)| = |0| = 0.

After the first update a becomes {11, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(11 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6) = |9| = 9.

After the second update a becomes {2, 2, 3, 4, 5} and it's optimal to choose j = 1, f(1) = |(2 - 2) - (2 - 3) + (3 - 4) - (4 - 5) + (5 - 6)| = |0| = 0.

After the third update a becomes {1, 1, 2, 3, 4} and it's optimal to choose j = 0, f(0) = |(1 - 1) - (1 - 2) + (2 - 3) - (3 - 4) + (4 - 5)| = |0| = 0.

Difficulty: interview

Starter code:

Solutions:
def read(): return list(map(int, input().split(' ')))
n, m, q = read()
aa = read()
bb = read()
reqs = [read() for _ in range(q)]

asum = 0
bsum = 0
for i, (a, b) in enumerate(zip(aa, bb)):
    asum += a if i % 2 == 0 else -a
    bsum += b if i % 2 == 0 else -b

bpos = [bsum]
for i in range(len(aa), len(bb)):
    b = bb[i]

    rempos = i - len(aa)
    bsum += b if i % 2 == 0 else -b
    bsum -= bb[rempos] if rempos % 2 == 0 else -bb[rempos]
    bpos += [bsum if rempos % 2 == 1 else -bsum]

bpos = sorted(set(bpos))

def closest(arr, value):
    l = 0
    r = len(arr)
    while l + 1 < r:
        m = (l + r) // 2
        if arr[m] <= value:
            l = m
        else:
            r = m
        
    res = arr[l]
    if l + 1 < len(arr) and abs(arr[l + 1] - value) < abs(arr[l] - value):
        res = arr[l + 1]
    return res

print(abs(asum - closest(bpos, asum)))
for req in reqs:
    l, r, x = req
    l -= 1
    if (r - l) % 2 != 0:
        asum += x if l % 2 == 0 else -x
    print(abs(asum - closest(bpos, asum)))