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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, m, q;
long long a[200200], b[200200], X, c[200200];
long long solve(long long C) {
  long long l = 0, r = m - n, ret = m - n + 1;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (c[mid] + C >= 0)
      ret = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  if (ret == m - n + 1)
    return abs(c[ret - 1] + C);
  else if (ret == 0)
    return abs(c[ret] + C);
  else
    return min(abs(c[ret - 1] + C), abs(c[ret] + C));
}
int main() {
  n = read(), m = read(), q = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= m; i++) b[i] = read();
  long long s = -1;
  for (long long j = 1, t = -1; j <= n; j++, t = -t) c[0] += t * b[j], s = t;
  for (long long i = 1, t = -1; i <= m - n; i++)
    c[i] = -(c[i - 1] - t * b[i]) + s * b[i + n];
  sort(c, c + 1 + m - n);
  for (long long i = 1, t = 1; i <= n; i++, t = -t) X += t * a[i];
  printf("%lld\n", solve(X));
  for (long long i = 1; i <= q; i++) {
    long long l = read(), r = read(), x = read();
    if ((r - l + 1) % 2 == 0)
      x = 0;
    else if (l & 1)
      X += x;
    else
      X -= x;
    printf("%lld\n", solve(X));
  }
}
