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>
const int MaxN = 2e5;
using namespace std;
int n, m, Q, tot;
long long sum_1, sum_2;
long long a[MaxN + 5], b[MaxN + 5], t[MaxN + 5];
long long A[MaxN + 5], B[MaxN + 5];
void Init() {
  sum_1 = sum_2 = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (i & 1)
      sum_1 = sum_1 + a[i];
    else
      sum_2 = sum_2 + a[i];
  }
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &b[i]);
    A[i] = A[i - 1] + ((i & 1) ? b[i] : 0);
    B[i] = B[i - 1] + ((i & 1) ? 0 : b[i]);
  }
  for (int i = 1; i <= m - n + 1; i++) {
    long long tot_2 = B[i + n - 1] - B[i - 1];
    long long tot_1 = A[i + n - 1] - A[i - 1];
    if (i & 1)
      t[i] = tot_2 - tot_1;
    else
      t[i] = tot_1 - tot_2;
  }
  tot = m - n + 1;
  sort(t + 1, t + tot + 1);
}
long long Query(long long tmp) {
  int l = 0, r = tot + 1;
  while (l < r - 1) {
    int mid = (l + r) >> 1;
    if (tmp + t[mid] <= 0)
      l = mid;
    else
      r = mid;
  }
  long long res = 1LL << 60;
  for (int i = max(1, r - 5); i <= min(tot, r + 5); i++) {
    res = min(res, abs(tmp + t[i]));
  }
  return res;
}
void Solve() {
  long long res = sum_1 - sum_2;
  printf("%lld\n", Query(res));
  for (int i = 1; i <= Q; i++) {
    int l, r;
    long long x;
    scanf("%d%d%lld", &l, &r, &x);
    if (((l & 1) ^ (r & 1)) == 0) {
      if (l & 1)
        res += x;
      else
        res -= x;
    }
    printf("%lld\n", Query(res));
  }
}
int main() {
  while (~scanf("%d%d%d", &n, &m, &Q)) {
    Init();
    Solve();
  }
  return 0;
}
