#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int maxn = 100005;
int n, m, q;
long long a[maxn], b[maxn], cur, f[maxn], pre[maxn];
vector<long long> v;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    cur += (i & 1) ? a[i] : -a[i];
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%I64d", &b[i]);
    pre[i] = pre[i - 1] + ((i & 1) ? -b[i] : b[i]);
  }
  for (int i = 0; i <= m - n; ++i) {
    v.push_back((i & 1) ? pre[i] - pre[i + n] : pre[i + n] - pre[i]);
  }
  sort(v.begin(), v.end());
  ++q;
  while (q--) {
    long long res = inf;
    int x = lower_bound(v.begin(), v.end(), -cur) - v.begin();
    if (x != (int)v.size()) res = abs(v[x] + cur);
    if (x > 0) res = min(res, abs(v[x - 1] + cur));
    printf("%I64d\n", res);
    if (q == 0) break;
    int l, r;
    long long t;
    scanf("%d%d%I64d", &l, &r, &t);
    cur += ((r - l + 1) % 2 == 1) ? ((l % 2 == 1) ? t : -t) : 0;
  }
  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 long long inf = 0x3f3f3f3f3f3f3f3f;
const int maxn = 100005;
int n, m, q;
long long a[maxn], b[maxn], cur, f[maxn], pre[maxn];
vector<long long> v;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    cur += (i & 1) ? a[i] : -a[i];
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%I64d", &b[i]);
    pre[i] = pre[i - 1] + ((i & 1) ? -b[i] : b[i]);
  }
  for (int i = 0; i <= m - n; ++i) {
    v.push_back((i & 1) ? pre[i] - pre[i + n] : pre[i + n] - pre[i]);
  }
  sort(v.begin(), v.end());
  ++q;
  while (q--) {
    long long res = inf;
    int x = lower_bound(v.begin(), v.end(), -cur) - v.begin();
    if (x != (int)v.size()) res = abs(v[x] + cur);
    if (x > 0) res = min(res, abs(v[x - 1] + cur));
    printf("%I64d\n", res);
    if (q == 0) break;
    int l, r;
    long long t;
    scanf("%d%d%I64d", &l, &r, &t);
    cur += ((r - l + 1) % 2 == 1) ? ((l % 2 == 1) ? t : -t) : 0;
  }
  return 0;
}

```