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>
const long long INF = 1e18 + 5;
const double PI = acos(-1.0);
const double EPS = 1e-9;
using namespace std;
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
template <class T>
void deb_array(T *arr, int length) {
  for (int i = 0; i < length; i++) {
    cout << arr[i] << ' ';
  }
  cout << '\n';
}
const int maxn = 1e5 + 5;
int n, m, q;
int a[maxn], b[maxn];
vector<long long> v;
long long s1, s2;
void solve() {
  int x = lower_bound(v.begin(), v.end(), s1) - v.begin();
  cout << min(abs(v[x] - s1), abs(v[x - 1] - s1)) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    s1 += (i % 2 == 0 ? a[i] : -a[i]);
  }
  for (int i = 0; i < m; i++) {
    cin >> b[i];
    if (i < n) {
      s2 += (i % 2 == 0 ? b[i] : -b[i]);
    }
  }
  v.push_back(INF);
  v.push_back(-INF);
  v.push_back(s2);
  for (int i = n; i < m; i++) {
    s2 -= b[i - n];
    s2 = -s2;
    s2 += (n % 2 == 1 ? b[i] : -b[i]);
    v.push_back(s2);
  }
  sort(v.begin(), v.end());
  cerr << "DEBUG: "
       << "v"
       << " = " << v << '\n';
  ;
  solve();
  for (int i = 0; i < q; i++) {
    int l, r, x;
    cin >> l >> r >> x;
    l--, r--;
    if (((r - l) % 2 == 0) && ((l % 2) == 1)) {
      s1 -= x;
    } else if (((r - l) % 2 == 0) && ((l % 2) == 0)) {
      s1 += x;
    }
    solve();
  }
  return 0;
}
