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, l, r;
long long a[100005], b[100005], c[100005], now, d;
set<long long> vis;
set<long long>::iterator it;
template <class T>
void chkmin(T &a, T b) {
  if (a > b) a = b;
}
long long calc(long long x) {
  it = vis.lower_bound(x);
  long long rtn = 1000000000000000000;
  if (it != vis.end()) chkmin(rtn, *it - x);
  if (it != vis.begin()) {
    it--;
    chkmin(rtn, x - *it);
  }
  return rtn;
}
int L, R;
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    now += (i & 1 ? -1 : 1) * a[i];
  }
  for (int i = 0; i < m; i++) {
    scanf("%I64d", &b[i]);
  }
  for (int i = 0; i < n; i++) {
    c[0] += (i & 1 ? -1 : 1) * b[i];
  }
  vis.insert(c[0]);
  for (int j = 1; j <= m - n; j++) {
    c[j] = -(c[j - 1] - b[j - 1]) + (n & 1 ? 1 : -1) * b[j + n - 1];
    vis.insert(c[j]);
  }
  printf("%I64d\n", calc(now));
  while (q--) {
    scanf("%d %d %I64d", &l, &r, &d);
    now += (r - l + 1) * d;
    if (L <= R) now -= 2 * (r / 2 - (l - 1) / 2) * d;
    printf("%I64d\n", calc(now));
  }
}
