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 n, m, q;
long long b[100000 + 1];
long long suma;
long long sums;
long long sum[100000 + 1];
long long sign(long long i) { return ((i & 1) ? 1 : (-1)); }
inline long long poz(long long a) { return (a > 0 ? a : (-a)); }
void readFile() {
  scanf("%lld%lld%lld", &n, &m, &q);
  long long i, nr;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &nr);
    suma += nr * sign(i);
  }
  long long sumb = 0;
  long long cr = 1;
  for (i = 1; i <= m; i++) {
    scanf("%lld", &b[i]);
    if (i > n) {
      sumb -= (-b[i - n]);
      sumb = (-sumb);
    }
    if (i <= n) cr = (-cr);
    sumb += cr * b[i];
    if (i >= n) sum[++sums] = sumb;
  }
}
long long elem(long long i) {
  if (i >= 1 && i <= sums) return sum[i];
  return sum[1];
}
void printRez() {
  long long i = 0;
  long long pas = (1 << 16);
  long long caut = -suma;
  while (pas != 0) {
    if (i + pas <= sums && sum[i + pas] <= caut) i += pas;
    pas >>= 1;
  }
  printf("%lld\n", min(min(poz(suma + elem(i)), poz(suma + elem(i + 1))),
                       poz(suma + elem(i - 1))));
}
void ansQues() {
  sort(sum + 1, sum + sums + 1);
  printRez();
  while (q > 0) {
    q--;
    long long st, dr, nr;
    scanf("%lld%lld%lld", &st, &dr, &nr);
    long long mns = (dr >> 1) - ((st - 1) >> 1);
    long long pls = (dr - st + 1) - mns;
    suma += 1LL * pls * nr;
    suma -= 1LL * mns * nr;
    printRez();
  }
}
int main() {
  readFile();
  ansQues();
  return 0;
}
