#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
long long sum[maxn], sum1[maxn], sum2[maxn], ans;
long long n, m, q, l, r, x, x1;
inline long long abs1(long long x) { return (x) > 0ll ? (x) : (-x); }
long long work(long long ans) {
  long long answer;
  int pos = lower_bound(sum, sum + m - n + 1, -ans) - sum;
  if (pos == 0)
    answer = abs(ans + sum[pos]);
  else if (pos == m - n + 1)
    answer = abs1(ans + sum[m - n]);
  else
    answer = min(abs1(ans + sum[pos - 1]), abs1(ans + sum[pos]));
  return answer;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &q);
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    if (i & 1)
      ans += x;
    else
      ans -= x;
  }
  for (long long i = 1; i <= m; i++) {
    scanf("%I64d", &x);
    if (i & 1)
      sum1[i] = sum1[i - 1] - x;
    else
      sum1[i] = sum1[i - 1] + x;
  }
  for (long long i = 0; i <= m - n; i++) {
    long long r = i + n;
    if (i & 1)
      sum[i] = -(sum1[r] - sum1[i]);
    else
      sum[i] = sum1[r] - sum1[i];
  }
  sort(sum, sum + m - n + 1);
  printf("%I64d\n", work(ans));
  for (long long i = 1; i <= q; i++) {
    scanf("%I64d%I64d%I64d", &l, &r, &x);
    if ((r - l + 1) & 1) {
      if (l & 1)
        ans += x;
      else
        ans -= x;
    }
    printf("%I64d\n", work(ans));
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int maxn = 100010;
long long sum[maxn], sum1[maxn], sum2[maxn], ans;
long long n, m, q, l, r, x, x1;
inline long long abs1(long long x) { return (x) > 0ll ? (x) : (-x); }
long long work(long long ans) {
  long long answer;
  int pos = lower_bound(sum, sum + m - n + 1, -ans) - sum;
  if (pos == 0)
    answer = abs(ans + sum[pos]);
  else if (pos == m - n + 1)
    answer = abs1(ans + sum[m - n]);
  else
    answer = min(abs1(ans + sum[pos - 1]), abs1(ans + sum[pos]));
  return answer;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &q);
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    if (i & 1)
      ans += x;
    else
      ans -= x;
  }
  for (long long i = 1; i <= m; i++) {
    scanf("%I64d", &x);
    if (i & 1)
      sum1[i] = sum1[i - 1] - x;
    else
      sum1[i] = sum1[i - 1] + x;
  }
  for (long long i = 0; i <= m - n; i++) {
    long long r = i + n;
    if (i & 1)
      sum[i] = -(sum1[r] - sum1[i]);
    else
      sum[i] = sum1[r] - sum1[i];
  }
  sort(sum, sum + m - n + 1);
  printf("%I64d\n", work(ans));
  for (long long i = 1; i <= q; i++) {
    scanf("%I64d%I64d%I64d", &l, &r, &x);
    if ((r - l + 1) & 1) {
      if (l & 1)
        ans += x;
      else
        ans -= x;
    }
    printf("%I64d\n", work(ans));
  }
  return 0;
}

```