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 MAXN = 1e5 + 5;
const long long INF = 1e18;
long long get(long long x) {
  if (x < 0) x = -x;
  return x;
}
long long a[MAXN], b[MAXN], c[MAXN], x, s[MAXN];
int n, m, q;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  int sg = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    x += a[i] * sg;
    sg = -sg;
  }
  for (int i = 1; i <= m; i++) scanf("%lld", b + i);
  sg = 1;
  for (int i = 1; i <= m; i++) {
    s[i] = s[i - 1] + sg * b[i];
    sg *= (-1);
  }
  for (int j = 0; j <= m - n; j++) {
    c[j] = s[j + n] - s[j];
    if (j % 2) c[j] = -c[j];
  }
  sort(c, c + m - n + 1);
  int pos = lower_bound(c, c + m - n + 1, x) - c;
  long long ans = INF;
  if (pos != m - n + 1) ans = min(ans, get(c[pos] - x));
  if (pos) ans = min(ans, get(c[pos - 1] - x));
  printf("%lld\n", ans);
  while (q--) {
    int l, r, alt;
    scanf("%d%d%d", &l, &r, &alt);
    if (r % 2 == l % 2) {
      if (l % 2)
        x += alt;
      else
        x -= alt;
    }
    pos = lower_bound(c, c + m - n + 1, x) - c;
    ans = INF;
    if (pos != m - n + 1) ans = min(ans, get(c[pos] - x));
    if (pos) ans = min(ans, get(c[pos - 1] - x));
    printf("%lld\n", ans);
  }
  return 0;
}
