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;
template <typename T1, typename T2>
inline void chkmin(T1 &first, T2 second) {
  if (first > second) first = second;
}
template <typename T1, typename T2>
inline void chkmax(T1 &first, T2 second) {
  if (first < second) first = second;
}
const string FILENAME = "input";
const int MAXN = 1 << 17;
int n, m, q;
long long suma, sumb;
long long a[MAXN], b[MAXN];
vector<long long> res;
void add(int i, int r) {
  if (i & 1) suma += r;
}
long long myabs(long long a) {
  if (a < 0) return -a;
  return a;
}
long long get() {
  int L = lower_bound(res.begin(), res.end(), -suma) - res.begin();
  L = max(0, L - 2);
  long long ans = 1e18;
  for (int i = L; i < res.size() && i < L + 3; ++i) {
    chkmin(ans, myabs(suma + res[i]));
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  srand(time(0));
  cin >> n >> m >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i & 1) {
      suma += a[i];
    } else {
      suma -= a[i];
    }
  }
  for (int i = 1; i <= m; ++i) {
    cin >> b[i];
    if (i & 1) {
      b[i] = -b[i] + b[i - 1];
    } else {
      b[i] = b[i] + b[i - 1];
    }
  }
  for (int j = 0; j <= m - n; ++j) {
    long long f = b[j + n] - b[j];
    if (j & 1) f = -f;
    res.push_back(f);
  }
  sort(res.begin(), res.end());
  cout << get() << endl;
  for (int i = 0; i < q; ++i) {
    int L, R, k;
    cin >> L >> R >> k;
    add(R, k);
    add(L - 1, -k);
    cout << get() << endl;
  }
}
