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, i;
long long a[100011], b[100011], sums[100011];
long long def, l, r, c;
vector<long long> ord;
long long abss(long long x) {
  if (x < 0) return -x;
  return x;
}
long long ask() {
  long long ans = 1LL << 60;
  auto it = lower_bound(ord.begin(), ord.end(), def);
  if (it != ord.end()) {
    ans = abss(def - *it);
  }
  if (it != ord.begin()) {
    it--;
    ans = min(ans, abss(def - *it));
  }
  return ans;
}
int main() {
  cin >> n >> m >> q;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (i & 1)
      def += a[i];
    else
      def -= a[i];
  }
  for (i = 1; i <= m; i++) {
    cin >> b[i];
    if (i % 2 == 0) b[i] *= -1;
    sums[i] = b[i] + sums[i - 1];
  }
  for (i = 1; i + n - 1 <= m; i++)
    ord.push_back((sums[i + n - 1] - sums[i - 1]) * (i % 2 == 1 ? 1 : -1));
  sort(ord.begin(), ord.end());
  printf("%lld\n", ask());
  for (i = 1; i <= q; i++) {
    scanf("%lld%lld%lld", &l, &r, &c);
    if ((r - l + 1) % 2 == 1) {
      if (l % 2 == 1)
        def += c;
      else
        def -= c;
    }
    printf("%lld\n", ask());
  }
  return 0;
}
