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 N = 1e5 + 10;
long long n, m, q, ans;
long long a[N], b[N], sum[N], f[N];
long long h(long long x) {
  long long l = 0, r = m - n, mid, cnt;
  cnt = 0x3f3f3f3f3f3f3f3f;
  while (l <= r) {
    mid = (l + r) / 2;
    if (x < f[mid] + ans)
      cnt = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  if (cnt == 0x3f3f3f3f3f3f3f3f) return cnt;
  return f[cnt];
}
long long lo(long long x) {
  long long l = 0, r = m - n, mid, cnt;
  cnt = 0x3f3f3f3f3f3f3f3f;
  while (l <= r) {
    mid = (l + r) / 2;
    if (x >= f[mid] + ans)
      cnt = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  if (cnt == 0x3f3f3f3f3f3f3f3f) return cnt;
  return f[cnt];
}
signed main() {
  scanf("%lld %lld %lld", &n, &m, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= m; i++) scanf("%lld", &b[i]);
  for (long long i = 1; i <= m; i++)
    if (i % 2) b[i] = -b[i];
  for (long long i = 1; i <= m; i++) sum[i] = sum[i - 1] + b[i];
  for (long long i = 1; i <= n; i++)
    if (!(i % 2)) ans -= 2 * a[i];
  for (long long i = 1; i <= n; i++) ans += a[i];
  for (long long i = 0; i <= m - n; i++) {
    if (i % 2 == 0)
      f[i] = ans + sum[n + i] - sum[i];
    else
      f[i] = ans + sum[i] - sum[n + i];
  }
  ans = 0;
  sort(f, f + m - n + 1);
  cout << min(abs(h(0)), abs(lo(0))) << '\n';
  for (long long i = 1; i <= q; i++) {
    long long l, r, k;
    scanf("%lld %lld %lld", &l, &r, &k);
    if (l % 2 && r % 2) ans += k;
    if (l % 2 == 0 && r % 2 == 0) ans -= k;
    cout << min(abs(h(0) + ans), abs(lo(0) + ans)) << '\n';
  }
  return 0;
}
