#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 100005;
const long long INF = 1000000009000000009;
int n, m, q;
int a[N], b[N];
long long s[N];
void solv() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
  for (int i = m; i >= 1; --i) {
    s[i] = s[i + 2] + b[i];
  }
  vector<long long> v;
  for (int i = 1; i <= m - n + 1; ++i) {
    int j1 = i + n, j2 = i + n;
    while (j1 % 2 != i % 2) ++j1;
    while (j2 % 2 != (i + 1) % 2) ++j2;
    v.push_back(s[i] - s[j1] - s[i + 1] + s[j2]);
  }
  sort(v.begin(), v.end());
  long long s = 0;
  for (int i = 1; i <= n; i += 2) s += a[i];
  for (int i = 2; i <= n; i += 2) s -= a[i];
  long long ans = INF;
  int j = upper_bound(v.begin(), v.end(), s) - v.begin();
  if (j < v.size()) ans = min(ans, v[j] - s);
  --j;
  if (j >= 0) ans = min(ans, s - v[j]);
  printf("%I64d\n", ans);
  while (q--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    if ((r - l + 1) % 2 == 1) {
      if (l % 2 == 1)
        s += x;
      else
        s -= x;
    }
    long long ans = INF;
    int j = upper_bound(v.begin(), v.end(), s) - v.begin();
    if (j < v.size()) ans = min(ans, v[j] - s);
    --j;
    if (j >= 0) ans = min(ans, s - v[j]);
    printf("%I64d\n", ans);
  }
}
int main() {
  solv();
  return 0;
}

### Prompt

Please create a solution in cpp 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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 100005;
const long long INF = 1000000009000000009;
int n, m, q;
int a[N], b[N];
long long s[N];
void solv() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
  for (int i = m; i >= 1; --i) {
    s[i] = s[i + 2] + b[i];
  }
  vector<long long> v;
  for (int i = 1; i <= m - n + 1; ++i) {
    int j1 = i + n, j2 = i + n;
    while (j1 % 2 != i % 2) ++j1;
    while (j2 % 2 != (i + 1) % 2) ++j2;
    v.push_back(s[i] - s[j1] - s[i + 1] + s[j2]);
  }
  sort(v.begin(), v.end());
  long long s = 0;
  for (int i = 1; i <= n; i += 2) s += a[i];
  for (int i = 2; i <= n; i += 2) s -= a[i];
  long long ans = INF;
  int j = upper_bound(v.begin(), v.end(), s) - v.begin();
  if (j < v.size()) ans = min(ans, v[j] - s);
  --j;
  if (j >= 0) ans = min(ans, s - v[j]);
  printf("%I64d\n", ans);
  while (q--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    if ((r - l + 1) % 2 == 1) {
      if (l % 2 == 1)
        s += x;
      else
        s -= x;
    }
    long long ans = INF;
    int j = upper_bound(v.begin(), v.end(), s) - v.begin();
    if (j < v.size()) ans = min(ans, v[j] - s);
    --j;
    if (j >= 0) ans = min(ans, s - v[j]);
    printf("%I64d\n", ans);
  }
}
int main() {
  solv();
  return 0;
}

```