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 sv(int p) {
  if (p % 2 == 0) return 1;
  return -1;
}
long long mod(long long a) {
  if (a >= 0) return a;
  return -1 * a;
}
long long min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
long long func(long long f[], long long d, int sz) {
  if (f[0] + d >= 0) return mod(d + f[0]);
  if (f[sz - 1] + d <= 0) return mod(d + f[sz - 1]);
  long long beg = 0, end = sz - 2;
  long long mid = (beg + end) / 2;
  while (beg <= end) {
    mid = (beg + end) / 2;
    if (f[mid + 1] + d < 0)
      beg = mid + 1;
    else if (f[mid] + d > 0)
      end = mid - 1;
    else if (f[mid] + d == 0)
      return 0;
    else if (f[mid + 1] + d == 0)
      return 0;
    else if (f[mid] + d < 0 && f[mid + 1] + d > 0)
      break;
  }
  return min(mod(f[mid] + d), mod(f[mid + 1] + d));
}
int main() {
  int n, m, q;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &q);
  long long a[n], b[m];
  int l[q], r[q];
  long long x[q], s = 0, d = 0;
  long long min = (0x3fffffffffffffff);
  ;
  long long f[m - n + 1];
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &a[i]);
  }
  for (int i = 0; i < m; ++i) {
    scanf("%I64d", &b[i]);
  }
  for (int i = 0; i <= m - n; ++i) {
    f[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    s += (sv(i) * a[i]);
  }
  for (int i = 0; i < n; ++i) {
    f[0] += (sv(i) * (a[i] - b[i]));
  }
  if (mod(f[0]) < min) min = mod(f[0]);
  for (int j = 1; j <= m - n; ++j) {
    f[j] = s + s - f[j - 1] - b[j - 1] + (sv(n) * b[j + n - 1]);
    if (mod(f[j]) < min) min = mod(f[j]);
  }
  printf("%I64d \n", min);
  sort(f, (f + (m - n) + 1));
  for (int i = 0; i < q; ++i) {
    cin >> l[i] >> r[i] >> x[i];
    if ((l[i] % 2) != (r[i] % 2))
      min = min;
    else if ((l[i] % 2) == 1) {
      d += x[i];
      min = func(f, d, m - n + 1);
    } else {
      d -= x[i];
      min = func(f, d, m - n + 1);
    }
    printf("%I64d \n", min);
  }
  return 0;
}
