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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD1 = 1e9 + 7, MOD2 = 1e9 + 9;
const int N = 1e5 + 10;
int n, m, q, x, b[N], k[] = {-1, 1};
long long sa = 0LL, tmp, sb[N];
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &x), sa += k[i % 2] * x;
  for (int i = 1; i <= m; i++)
    scanf("%d", &b[i]), sb[1] += (i <= n ? k[i % 2] * b[i] : 0);
  for (int i = 2; i + n - 1 <= m; i++)
    sb[i] = -(sb[i - 1] - b[i - 1]) + k[n % 2] * b[i + n - 1];
  sort(sb + 1, sb + m - n + 2);
  sb[0] = -LINF, sb[m - n + 2] = LINF;
  long long ans = LINF;
  tmp = lower_bound(sb + 1, sb + m - n + 1, sa) - sb;
  ans = min(ans, min(abs(sa - sb[tmp]),
                     min(abs(sa - sb[tmp - 1]), abs(sa - sb[tmp + 1]))));
  printf("%lld\n", ans);
  int l, r, x;
  while (q--) {
    scanf("%d %d %d", &l, &r, &x);
    if ((r - l + 1) % 2) sa += k[l % 2] * x;
    tmp = lower_bound(sb + 1, sb + m - n + 1, sa) - sb;
    ans = LINF;
    ans = min(ans, min(abs(sa - sb[tmp]),
                       min(abs(sa - sb[tmp - 1]), abs(sa - sb[tmp + 1]))));
    printf("%lld\n", ans);
  }
}
