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 int n, m, q;
long long int a[100005], b[100005];
set<long long int> s;
long long int ans = 0;
void eval(long long int x) {
  ans = 1e18;
  auto h = s.lower_bound(-x);
  if (h != s.end()) ans = min(ans, abs(x + *h));
  if (h != s.begin()) {
    h--;
    ans = min(ans, abs(x + *h));
  }
}
int main() {
  scanf("%lld", &n);
  scanf("%lld", &m);
  scanf("%lld", &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= m; i++) scanf("%lld", &b[i]);
  for (int i = 1; i <= n; i++)
    if (i % 2)
      a[i] = a[i - 1] + a[i];
    else
      a[i] = a[i - 1] - a[i];
  for (int i = 1; i <= m; i++)
    if (i % 2)
      b[i] = b[i - 1] + b[i];
    else
      b[i] = b[i - 1] - b[i];
  long long int sum = a[n];
  for (int j = 1; j <= m - n + 1; j++) {
    long long int x = b[j + n - 1] - b[j - 1];
    if (j % 2 == 1) x *= -1;
    s.insert(x);
  }
  auto h = s.begin();
  eval(sum);
  printf("%lld\n", ans);
  for (int i = 1; i <= q; i++) {
    long long int l, r, x;
    scanf("%lld", &l);
    scanf("%lld", &r);
    scanf("%lld", &x);
    if ((r - l + 1) % 2 != 0) {
      if (l % 2)
        sum += x;
      else
        sum -= x;
    }
    eval(sum);
    printf("%lld\n", ans);
  }
  return 0;
}
