#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 3;
int n, m, q;
long long int a[MAXN], b[MAXN], sumB, sumA;
vector<long long int> V;
void bs() {
  if (V.back() <= sumA || V[0] >= sumA) {
    printf("%lld\n", min(abs(sumA - V.back()), abs(sumA - V[0])));
    return;
  }
  int hi = (int)V.size() - 1, lo = 0, mid;
  long long int diff, ans = abs(sumA - V[0]);
  while (hi >= lo && ans) {
    mid = (hi + lo) >> 1;
    diff = sumA - V[mid];
    if (diff > 0) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
    ans = min(ans, abs(diff));
  }
  printf("%lld\n", ans);
}
int main() {
  sumB = sumA = 0;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    if (i & 1)
      sumA += a[i];
    else
      sumA -= a[i];
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%lld", b + i);
    if (i <= n) {
      if (i & 1)
        sumB += b[i];
      else
        sumB -= b[i];
    }
  }
  V.push_back(sumB);
  for (int j = n + 1; j <= m; ++j) {
    sumB -= b[j - n];
    sumB *= -1;
    if (n & 1)
      sumB += b[j];
    else
      sumB -= b[j];
    V.push_back(sumB);
  }
  sort(V.begin(), V.end());
  bs();
  while (q--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    if ((r - l + 1) & 1) {
      if (l & 1)
        sumA += x;
      else
        sumA -= x;
    }
    bs();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 1e5 + 3;
int n, m, q;
long long int a[MAXN], b[MAXN], sumB, sumA;
vector<long long int> V;
void bs() {
  if (V.back() <= sumA || V[0] >= sumA) {
    printf("%lld\n", min(abs(sumA - V.back()), abs(sumA - V[0])));
    return;
  }
  int hi = (int)V.size() - 1, lo = 0, mid;
  long long int diff, ans = abs(sumA - V[0]);
  while (hi >= lo && ans) {
    mid = (hi + lo) >> 1;
    diff = sumA - V[mid];
    if (diff > 0) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
    ans = min(ans, abs(diff));
  }
  printf("%lld\n", ans);
}
int main() {
  sumB = sumA = 0;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    if (i & 1)
      sumA += a[i];
    else
      sumA -= a[i];
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%lld", b + i);
    if (i <= n) {
      if (i & 1)
        sumB += b[i];
      else
        sumB -= b[i];
    }
  }
  V.push_back(sumB);
  for (int j = n + 1; j <= m; ++j) {
    sumB -= b[j - n];
    sumB *= -1;
    if (n & 1)
      sumB += b[j];
    else
      sumB -= b[j];
    V.push_back(sumB);
  }
  sort(V.begin(), V.end());
  bs();
  while (q--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    if ((r - l + 1) & 1) {
      if (l & 1)
        sumA += x;
      else
        sumA -= x;
    }
    bs();
  }
  return 0;
}

```