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;
const int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
const long double pi = acos(-1);
const int MOD = 1e9 + 7;
const int N = 1e3 + 10;
long long int sa;
vector<long long int> sbp, sbn;
int n, m, q;
bool tem0;
long long int b[100050];
void pre() {
  cin >> n >> m >> q;
  long long int s1 = 1, s2 = -1;
  for (int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    sa += s1 * x;
    swap(s1, s2);
  }
  s1 = 1, s2 = -1;
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
    b[i] *= s1;
    b[i] += b[i - 1];
    swap(s1, s2);
  }
  s1 = 1, s2 = -1;
  for (int i = 0; i <= m - n; i++) {
    int j = i + n;
    long long int sum = b[j] - b[i];
    sum *= s1;
    if (sum > 0) {
      sbp.push_back(sum);
    } else {
      sbn.push_back(sum);
    }
    swap(s1, s2);
  }
  sort(sbn.begin(), sbn.end());
  sort(sbp.begin(), sbp.end());
}
int bb(int e, int d, vector<long long int> &v) {
  if (e >= d) return e;
  int m = (e + d + 1) / 2;
  if (v[m] > sa)
    return bb(e, m - 1, v);
  else
    return bb(m, d, v);
}
long long int get() {
  long long int x = LINF;
  if (sbp.size() > 0) {
    int id = bb(0, sbp.size() - 1, sbp);
    x = min(x, abs(sa - sbp[id]));
    id++;
    if (id != sbp.size()) x = min(x, abs(sa - sbp[id]));
  }
  if (sbn.size() > 0) {
    int id = bb(0, sbn.size() - 1, sbn);
    x = min(x, abs(sa - sbn[id]));
    id++;
    if (id != sbn.size()) x = min(x, abs(sa - sbn[id]));
  }
  return x;
}
void vai(long long int e, long long int d, long long int x) {
  if (e % 2 == d % 2) {
    if (e % 2 == 1)
      sa += x;
    else
      sa -= x;
  }
}
int main() {
  ios::sync_with_stdio(false);
  ;
  pre();
  cout << get() << endl;
  while (q--) {
    long long int e, d, x;
    cin >> e >> d >> x;
    vai(e, d, x);
    cout << get() << endl;
  }
  return 0;
}
