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E. Mahmoud and Ehab and the function
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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, .

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.

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 a 1, a 2, ..., a n. ( - 109 ≤ a i ≤ 109) — elements of a.

The third line contains m integers b 1, b 2, ..., b m. ( - 109 ≤ b i ≤ 109) — 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,  - 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 31 2 3 4 51 2 3 4 5 61 1 101 1 -91 5 -1
Output
0900
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.