#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5;
long long prefixA = 0;
long long prefixB[maxn + 10];
long long sumB[maxn + 10];
long long b[maxn + 10];
void solve() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    if (i & 1) {
      prefixA += x;
    } else
      prefixA -= x;
  }
  sumB[0] = 0;
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &b[i]);
    if (i & 1) {
      sumB[i] = sumB[i - 1] + b[i];
    } else {
      sumB[i] = sumB[i - 1] - b[i];
    }
  }
  for (int i = 1; i <= m - n + 1; i++) {
    if (!(i & 1)) {
      prefixB[i - 1] = sumB[i - 1 + n] - sumB[i - 1];
    } else {
      prefixB[i - 1] = sumB[i - 1] - sumB[i - 1 + n];
    }
  }
  sort(prefixB, prefixB + m - n + 1);
  int pos = lower_bound(prefixB, prefixB + m - n + 1, -prefixA) - prefixB;
  pos = min(pos, m - n);
  printf("%lld\n", min(min(abs(prefixA + prefixB[max(pos - 1, 0)]),
                           abs(prefixA + prefixB[pos])),
                       abs(prefixA + prefixB[min(pos + 1, m - n)])));
  while (q--) {
    long long l, r, x;
    scanf("%lld%lld%lld", &l, &r, &x);
    if ((r - l + 1) & 1) {
      if (l & 1) {
        prefixA += x;
      } else {
        prefixA -= x;
      }
    }
    pos = lower_bound(prefixB, prefixB + m - n + 1, -prefixA) - prefixB;
    pos = min(pos, m - n);
    printf("%lld\n", min(min(abs(prefixA + prefixB[max(pos - 1, 0)]),
                             abs(prefixA + prefixB[pos])),
                         abs(prefixA + prefixB[min(pos + 1, m - n)])));
  }
}
int main() {
  solve();
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5;
long long prefixA = 0;
long long prefixB[maxn + 10];
long long sumB[maxn + 10];
long long b[maxn + 10];
void solve() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    if (i & 1) {
      prefixA += x;
    } else
      prefixA -= x;
  }
  sumB[0] = 0;
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &b[i]);
    if (i & 1) {
      sumB[i] = sumB[i - 1] + b[i];
    } else {
      sumB[i] = sumB[i - 1] - b[i];
    }
  }
  for (int i = 1; i <= m - n + 1; i++) {
    if (!(i & 1)) {
      prefixB[i - 1] = sumB[i - 1 + n] - sumB[i - 1];
    } else {
      prefixB[i - 1] = sumB[i - 1] - sumB[i - 1 + n];
    }
  }
  sort(prefixB, prefixB + m - n + 1);
  int pos = lower_bound(prefixB, prefixB + m - n + 1, -prefixA) - prefixB;
  pos = min(pos, m - n);
  printf("%lld\n", min(min(abs(prefixA + prefixB[max(pos - 1, 0)]),
                           abs(prefixA + prefixB[pos])),
                       abs(prefixA + prefixB[min(pos + 1, m - n)])));
  while (q--) {
    long long l, r, x;
    scanf("%lld%lld%lld", &l, &r, &x);
    if ((r - l + 1) & 1) {
      if (l & 1) {
        prefixA += x;
      } else {
        prefixA -= x;
      }
    }
    pos = lower_bound(prefixB, prefixB + m - n + 1, -prefixA) - prefixB;
    pos = min(pos, m - n);
    printf("%lld\n", min(min(abs(prefixA + prefixB[max(pos - 1, 0)]),
                             abs(prefixA + prefixB[pos])),
                         abs(prefixA + prefixB[min(pos + 1, m - n)])));
  }
}
int main() {
  solve();
  return 0;
}

```