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;
int num[2][100005], cnt = 0;
long long ans = 0;
long long sum = 0, nums[100005];
long long ask1() {
  if (nums[cnt] <= ans) return ans - nums[cnt];
  if (nums[1] >= ans) return nums[1] - ans;
  int l = 1, r = cnt, mid;
  while (l != r) {
    mid = (l + r) / 2;
    if (nums[mid] >= ans)
      r = mid;
    else
      l = mid + 1;
  }
  return nums[l] - ans;
}
long long ask2() {
  if (nums[cnt] <= ans) return ans - nums[cnt];
  if (nums[1] >= ans) return nums[1] - ans;
  int l = 1, r = cnt, mid;
  while (l != r) {
    mid = (l + r + 1) / 2;
    if (nums[mid] <= ans)
      l = mid;
    else
      r = mid - 1;
  }
  return ans - nums[l];
}
int main() {
  int n, m, k;
  int l, r, now;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &num[0][i]);
  for (int i = 1; i <= m; i++) scanf("%d", &num[1][i]);
  for (int i = 1; i <= n; i++) {
    if (i % 2) {
      ans += num[0][i];
      sum -= num[1][i];
    } else {
      ans -= num[0][i];
      sum += num[1][i];
    }
  }
  cnt++;
  nums[cnt] = sum;
  for (int i = 1; i <= m - n; i++) {
    sum = -sum - num[1][i];
    if (n % 2)
      sum -= num[1][i + n];
    else
      sum += num[1][i + n];
    cnt++;
    nums[cnt] = sum;
  }
  for (int i = 1; i <= m - n + 1; i++) nums[i] = -nums[i];
  sort(nums + 1, nums + 1 + cnt);
  while (k--) {
    printf("%lld\n", min(ask1(), ask2()));
    scanf("%d%d%d", &l, &r, &now);
    if (l % 2 == r % 2) {
      if (l % 2)
        ans += now;
      else
        ans -= now;
    }
  }
  printf("%lld\n", min(ask1(), ask2()));
  return 0;
}
