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;
long long a_sum, b_sum, b_even_sum[100005], b_odd_sum[100005];
int n, m, t, q;
vector<long long> b;
long long get(long long x) {
  vector<long long>::iterator it = lower_bound(b.begin(), b.end(), x);
  if (it == b.end()) it--;
  long long ret = abs((*it) - x);
  if (it != b.begin()) it--;
  return min(ret, abs((*it) - x));
}
int main() {
  int i, j;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &t);
    if (i % 2)
      a_sum += t;
    else
      a_sum -= t;
  }
  for (i = 1; i <= m; i++) {
    scanf("%d", &t);
    if (i % 2) {
      b_odd_sum[i] = b_odd_sum[i - 1] + t;
      b_even_sum[i] = b_even_sum[i - 1];
    } else {
      b_even_sum[i] = b_even_sum[i - 1] + t;
      b_odd_sum[i] = b_odd_sum[i - 1];
    }
  }
  for (j = 0; j <= m - n; j++) {
    int l = 1 + j, r = n + j;
    if (l % 2) {
      b_sum = (b_odd_sum[r] - b_odd_sum[l - 1]) -
              (b_even_sum[r] - b_even_sum[l - 1]);
    } else {
      b_sum = (b_even_sum[r] - b_even_sum[l - 1]) -
              (b_odd_sum[r] - b_odd_sum[l - 1]);
    }
    b.push_back(b_sum);
  }
  sort(b.begin(), b.end());
  printf("%lld\n", get(a_sum));
  for (i = 0; i < q; i++) {
    int l, r, x;
    scanf("%d", &l);
    scanf("%d", &r);
    scanf("%d", &x);
    if ((r - l + 1) % 2 == 0) {
    } else if (l % 2) {
      a_sum += x;
    } else {
      a_sum -= x;
    }
    printf("%lld\n", get(a_sum));
  }
  return 0;
}
