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;
int n, m, q;
long long F[100000];
long long B[100050];
long long lower_find(long long x) {
  int L = 0, R = m - n;
  while (R > L) {
    int mid = (R + L + 1) / 2;
    if (F[mid] == x)
      return 0;
    else if (F[mid] > x)
      R = mid - 1;
    else
      L = mid;
  }
  long long x1, x2;
  x1 = abs(F[L] - x);
  if (L < m - n)
    x2 = abs(F[L + 1] - x);
  else
    x2 = 9223372036854775807;
  return x1 < x2 ? x1 : x2;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  long long f = 1, a, b;
  long long suma = 0, sumb = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a);
    suma += f * a;
    f = -f;
  }
  f = -1;
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &B[i]);
    sumb += B[i] * f;
    f = -f;
    if (i >= n) {
      int t = (i - n) % 2 ? -1 : 1;
      F[i - n] = suma + sumb * t;
      sumb += B[i - n + 1] * t;
    }
  }
  sort(F, F + m - n + 1);
  int l, r;
  long long x, cur = 0;
  printf("%lld\n", lower_find(0));
  for (int i = 0; i < q; i++) {
    scanf("%d%d%lld", &l, &r, &x);
    if ((r - l) % 2 == 0)
      if (l % 2 == 1)
        cur += x;
      else
        cur -= x;
    printf("%lld\n", lower_find(-cur));
  }
  return 0;
}
