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 + 10;
int N, M, Q;
long long sa, pb[MAXN];
set<long long> st;
long long off;
long long calc() {
  long long res = min(llabs(*st.begin() + off), llabs(*st.rbegin() + off));
  auto it = st.lower_bound(-off);
  if (it != st.end()) {
    res = min(res, (*it) + off);
  }
  if (it != st.begin()) {
    it--;
    res = min(res, -((*it) + off));
  }
  return res;
}
int main() {
  scanf("%d %d %d", &N, &M, &Q);
  for (int i = 1; i <= N; i++) {
    long long a;
    scanf("%lld", &a);
    if (i % 2 == 0) {
      a *= -1;
    }
    sa += a;
  }
  for (int i = 1; i <= M; i++) {
    scanf("%lld", &pb[i]);
    if (i % 2) {
      pb[i] *= -1;
    }
    pb[i] += pb[i - 1];
  }
  for (int i = 0; i <= M - N; i++) {
    long long sb = pb[i + N] - pb[i];
    if (i % 2) {
      sb *= -1;
    }
    st.insert(sa + sb);
  }
  printf("%lld\n", calc());
  for (int qi = 1; qi <= Q; qi++) {
    int lt, rt, x;
    scanf("%d %d %d", &lt, &rt, &x);
    if (lt % 2 == rt % 2) {
      if (lt % 2) {
        off += x;
      } else {
        off -= x;
      }
    }
    printf("%lld\n", calc());
  }
}
