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;
template <typename TP>
inline bool rd(TP& r) {
  r = 0;
  char tmp = getchar();
  while (tmp < '0' || tmp > '9') {
    if (tmp == EOF) return 0;
    tmp = getchar();
  }
  while ('0' <= tmp && tmp <= '9') {
    r = (r << 3) + (r << 1) + tmp - '0';
    tmp = getchar();
  }
  return 1;
}
long long ab(long long num) {
  if (num >= 0) return num;
  return -num;
}
int n, m, q;
long long a[100100], b[100100];
long long suma;
long long g[100100];
int main() {
  rd(n), rd(m), rd(q);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d\n", &a[i]);
    if (i & 1)
      suma += a[i];
    else
      suma -= a[i];
  }
  for (int i = 1; i <= m; i++) scanf("%I64d\n", &b[i]);
  for (int i = 1; i <= n; i++)
    if (i & 1)
      g[0] += b[i];
    else
      g[0] -= b[i];
  for (int j = 1; j <= m - n; j++) {
    g[j] = -(g[j - 1] - b[j]);
    if (n & 1)
      g[j] += b[n + j];
    else
      g[j] -= b[n + j];
  }
  sort(g, g + m - n + 1);
  int l, r, po;
  long long ans = 0;
  po = lower_bound(g, g + m - n + 1, suma) - g;
  ans = 1000000000000000000;
  if (po >= 0 && g[po] == suma) ans = 0;
  if (po != 0) {
    po--;
    ans = min(ans, ab(suma - g[po]));
  }
  po = upper_bound(g, g + m - n + 1, suma) - g;
  if (po <= m - n) ans = min(ans, ab(suma - g[po]));
  printf("%I64d\n", ans);
  long long x;
  for (int i = 1; i <= q; i++) {
    rd(l), rd(r);
    scanf("%I64d\n", &x);
    if ((r - l + 1) & 1) {
      if (r & 1)
        suma += x;
      else
        suma -= x;
    }
    po = lower_bound(g, g + m - n + 1, suma) - g;
    ans = 1000000000000000000;
    if (po >= 0 && g[po] == suma) ans = 0;
    if (po != 0) {
      po--;
      ans = min(ans, ab(suma - g[po]));
    }
    po = upper_bound(g, g + m - n + 1, suma) - g;
    if (po <= m - n) ans = min(ans, ab(suma - g[po]));
    printf("%I64d\n", ans);
  }
  return 0;
}
