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;
long long INF = 1e18;
const int N = 1e5 + 111;
long long a[N], b[N], c[N];
int main() {
  long long n, m, q, pos, sum = 0, tmp;
  cin >> n >> m >> q;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    i % 2 == 1 ? sum += a[i] : sum -= a[i];
  }
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &tmp);
    if (i % 2 == 0) tmp *= -1;
    b[i] = b[i - 1] + tmp;
  }
  for (int i = 1; i <= m - n + 1; i++) {
    if (i % 2)
      c[i] = (b[i - 1] - b[i + n - 1]);
    else
      c[i] = (b[i + n - 1] - b[i - 1]);
  }
  sort(c + 1, c + 1 + m - n + 1);
  long long ans = INF, x, y, z;
  pos = lower_bound(c + 1, c + m - n + 1, -sum) - c;
  if (pos == 1)
    ans = abs(sum + c[pos]);
  else
    ans = min(abs(sum + c[pos]), abs(sum + c[pos - 1]));
  printf("%lld\n", ans);
  for (int k = 1; k <= q; k++) {
    ans = INF;
    scanf("%lld %lld %lld", &x, &y, &z);
    if (y == x) {
      y % 2 == 1 ? sum += z : sum -= z;
    } else if ((y - x + 1) % 2) {
      x % 2 == 1 ? sum += z : sum -= z;
    }
    pos = lower_bound(c + 1, c + m - n + 1, -sum) - c;
    if (pos == 1)
      ans = abs(sum + c[pos]);
    else
      ans = min(abs(sum + c[pos]), abs(sum + c[pos - 1]));
    printf("%lld\n", ans);
  }
  return 0;
}
