#include <bits/stdc++.h>
using namespace std;
long long n, m, q, a[100010], b[100010], s[100010], c[100010];
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &q);
  long long i, j;
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (i = 1; i <= m; i++) scanf("%I64d", &b[i]);
  for (i = 1, j = 1, s[0] = 0; i <= n; i++, j *= -1) s[0] += b[i] * j;
  for (i = 1; i <= m - n; i++)
    s[i] = -(s[i - 1] - b[i]) + b[n + i] * (n % 2 ? 1 : -1);
  for (a[0] = 0, i = 1, j = 1; i <= n; i++, j *= -1) a[0] += a[i] * j;
  for (i = 0; i <= m - n; i++) c[i] = a[0] - s[i];
  sort(c, c + m - n + 1);
  long long l, r, x, t = 0, lb, rb, mid, ans;
  for (i = 0; i <= q; i++) {
    if (i) {
      scanf("%I64d%I64d%I64d", &l, &r, &x);
      if (l % 2 && r % 2)
        t += x;
      else if (!(l % 2) && !(r % 2))
        t -= x;
    }
    lb = 0;
    rb = m - n;
    while (lb < rb) {
      mid = (lb + rb) / 2;
      if (c[mid] + t >= 0)
        rb = mid;
      else
        lb = mid + 1;
    }
    ans = abs(c[lb] + t);
    if (lb) ans = min(ans, abs(c[lb - 1] + t));
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, q, a[100010], b[100010], s[100010], c[100010];
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &q);
  long long i, j;
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (i = 1; i <= m; i++) scanf("%I64d", &b[i]);
  for (i = 1, j = 1, s[0] = 0; i <= n; i++, j *= -1) s[0] += b[i] * j;
  for (i = 1; i <= m - n; i++)
    s[i] = -(s[i - 1] - b[i]) + b[n + i] * (n % 2 ? 1 : -1);
  for (a[0] = 0, i = 1, j = 1; i <= n; i++, j *= -1) a[0] += a[i] * j;
  for (i = 0; i <= m - n; i++) c[i] = a[0] - s[i];
  sort(c, c + m - n + 1);
  long long l, r, x, t = 0, lb, rb, mid, ans;
  for (i = 0; i <= q; i++) {
    if (i) {
      scanf("%I64d%I64d%I64d", &l, &r, &x);
      if (l % 2 && r % 2)
        t += x;
      else if (!(l % 2) && !(r % 2))
        t -= x;
    }
    lb = 0;
    rb = m - n;
    while (lb < rb) {
      mid = (lb + rb) / 2;
      if (c[mid] + t >= 0)
        rb = mid;
      else
        lb = mid + 1;
    }
    ans = abs(c[lb] + t);
    if (lb) ans = min(ans, abs(c[lb - 1] + t));
    printf("%I64d\n", ans);
  }
  return 0;
}

```