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, ci = ai - bi + j. Then f(j) = |c1 - c2 + c3 - c4... cn|. More formally, <image>. 

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 xi to all elements in a in range [li;ri] i.e. they should add xi to ali, ali + 1, ... , ari 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 ≤ 105, 1 ≤ q ≤ 105) — number of elements in a, number of elements in b and number of queries, respectively.

The second line contains n integers a1, a2, ..., an. ( - 109 ≤ ai ≤ 109) — elements of a.

The third line contains m integers b1, b2, ..., bm. ( - 109 ≤ bi ≤ 109) — elements of b.

Then q lines follow describing the queries. Each of them contains three integers li ri xi (1 ≤ li ≤ ri ≤ n,  - 109 ≤ x ≤ 109) — 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.
import sys
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]
ii = 0

n = inp[ii]
ii += 1
m = inp[ii]
ii += 1
q = inp[ii]
ii += 1

A = inp[ii:ii+n]
ii += n

Asum = sum(A[::2]) - sum(A[1::2])

B = inp[ii:ii+m]
ii += m

B = [B[i]*(1 - 2*(i&1)) for i in range(m)]

Bsum = sum(B[:n])
Bvals = [Bsum]
for j in range(1, m - n + 1):
    Bsum -= B[j - 1]
    Bsum += B[j + n - 1]
    Bvals.append(Bsum * (1 - 2*(j&1)))

Bvals.sort()

out = []

def solve():
    a = 0
    b = len(Bvals)
    while 1 + a - b:
        c = a + b >> 1
        if Bvals[c] <= Asum:
            a = c
        else:
            b = c
    if a + 1 < len(Bvals):
        return min(abs(Asum - Bvals[a]), abs(Asum - Bvals[a + 1]))
    return abs(Asum - Bvals[a])

out.append(solve())

for _ in range(q):
    l = inp[ii] - 1
    ii += 1
    r = inp[ii]
    ii += 1

    x = inp[ii]
    ii += 1

    fac = 0
    if l&1:
        fac -= 1
        l += 1
    if (r - l)&1:
        fac += 1
    Asum += fac * x

    out.append(solve())
print '\n'.join(str(x) for x in out)
