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;
set<long long> s;
set<long long>::iterator it;
int n, m, x, y, z, k, w, l, r;
int A[100005], B[100005];
long long sum, tmp;
long long solve(long long sum) {
  long long t1 = -1, t2 = -1;
  it = s.lower_bound(sum);
  if (it != s.end()) {
    t1 = abs(*it - sum);
  }
  if (it != s.begin()) {
    --it;
    t2 = abs(*it - sum);
  }
  if (t1 == -1)
    return t2;
  else if (t2 == -1)
    return t1;
  else
    return min(t1, t2);
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = (0); i < (n); i++) {
    scanf("%lld", &A[i]);
    if (i % 2 == 0)
      sum += A[i];
    else
      sum -= A[i];
  }
  for (int i = (0); i < (m); i++) scanf("%lld", &B[i]);
  for (int i = (0); i < (m); i++) {
    if (i % 2 == 0)
      tmp -= B[i];
    else
      tmp += B[i];
    if (i >= n - 1) {
      if (i >= n) {
        if ((i - n) % 2 == 0)
          tmp += B[i - n];
        else
          tmp -= B[i - n];
      }
      if ((i - (n - 1)) % 2 == 0)
        s.insert(tmp);
      else
        s.insert(-tmp);
    }
  }
  printf("%lld\n", solve(-sum));
  while (k--) {
    scanf("%d %d %lld", &l, &r, &tmp);
    if (r % 2 == 0 && l % 2 == 0)
      sum -= tmp;
    else if (r % 2 == 1 && l % 2 == 1)
      sum += tmp;
    printf("%lld\n", solve(-sum));
  }
  return 0;
}
