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 sign(int x) {
  if (x % 2) return -1;
  return 1;
}
long long absll(long long x) {
  if (x > 0) return x;
  return -x;
}
set<long long> s;
int b[100005];
long long ans(long long x) {
  set<long long>::iterator it = s.lower_bound(x);
  if (it == s.end()) it--;
  long long ret = absll((*it) - x);
  if (it != s.begin()) it--;
  return min(ret, absll((*it) - x));
}
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  long long sumA = 0;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    sumA += a * sign(i);
  }
  long long cur = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    cur += sign(i + 1) * b[i];
  }
  s.insert(cur);
  for (int i = n; i < m; i++) {
    scanf("%d", &b[i]);
    cur += b[i - n];
    cur = -cur;
    cur += sign(n) * b[i];
    s.insert(cur);
  }
  printf("%I64d\n", ans(-sumA));
  while (q--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    if ((r - l) % 2 == 0) {
      if (l % 2)
        sumA += x;
      else
        sumA -= x;
    }
    printf("%I64d\n", ans(-sumA));
  }
}
