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.
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int b[SIZE];
void answer(vector<long long>& d, long long v) {
  int it = lower_bound((d).begin(), (d).end(), -v) - d.begin();
  long long an = 1e18;
  if (it != ((int)(d).size())) an = min(an, abs(d[it] + v));
  it--;
  if (it >= 0) an = min(an, abs(d[it] + v));
  printf("%I64d\n", an);
}
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  long long base = 0;
  for (int i = 0; i < (n); ++i) {
    int(x);
    scanf("%d", &x);
    if (i & 1)
      base -= x;
    else
      base += x;
  }
  long long now = 0;
  vector<long long> d;
  for (int i = (1); i < (m + 1); ++i) {
    scanf("%d", &(b[i]));
    if (i & 1)
      now -= b[i];
    else
      now += b[i];
    if (i >= n) {
      if ((i - n) % 2 == 0)
        d.push_back(now);
      else
        d.push_back(-now);
      if ((i - n + 1) & 1)
        now += b[i - n + 1];
      else
        now -= b[i - n + 1];
    }
  }
  sort((d).begin(), (d).end());
  answer(d, base);
  while (q--) {
    int x, y, v;
    scanf("%d%d%d", &x, &y, &v);
    if ((y - x + 1) % 2 == 1) {
      if (x & 1)
        base += v;
      else
        base -= v;
    }
    answer(d, base);
  }
  return 0;
}
