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;
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline long long minize(long long& a, long long b) {
  return a = (a > b ? b : a);
}
inline long long maxize(long long& a, long long b) {
  return a = (a < b ? b : a);
}
inline string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
double const eps = 1e-6;
long long const Base = 1e9 + 7, oo = 1e17, MAXN = 1e5;
vector<long long> arrvalb;
long long arra[MAXN + 5], arrb[MAXN + 5], sumb[MAXN + 5];
long long Solves() {
  long long n, m, k, cnt = 0, ans = 0, x, y, q, c, sum = 0, v, t, vala = 0, l,
                     r;
  cin >> n >> m;
  cin >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> arra[i];
    if (i & 1)
      vala += arra[i];
    else
      vala -= arra[i];
  }
  for (long long i = 1; i <= m; i++) {
    cin >> arrb[i];
    if (i > 1)
      sumb[i] = arrb[i] + sumb[i - 2];
    else
      sumb[i] = arrb[i];
  }
  for (long long i = n; i <= m; i++) {
    l = i - n + 1;
    r = i;
    if (l > 1)
      x = -(sumb[l + 2 * ((r - l + 2) / 2 - 1)] - sumb[l - 2]);
    else
      x = -sumb[l + 2 * ((r - l + 2) / 2 - 1)];
    if (r - l != 0) {
      l++;
      x += sumb[l + 2 * ((r - l + 2) / 2 - 1)] - sumb[l - 2];
    }
    arrvalb.push_back(x);
  }
  arrvalb.push_back(-oo);
  arrvalb.push_back(oo);
  sort((arrvalb).begin(), (arrvalb).end());
  r = upper_bound((arrvalb).begin(), (arrvalb).end(), -vala) - arrvalb.begin();
  l = lower_bound((arrvalb).begin(), (arrvalb).end(), -vala) - arrvalb.begin();
  if (arrvalb[l] != -vala) l--;
  cout << (min(abs(vala + arrvalb[l]), abs(vala + arrvalb[r]))) << endl;
  while (q--) {
    cin >> l >> r >> x;
    if ((r - l + 1) & 1) {
      if (l & 1)
        vala += x;
      else
        vala -= x;
    }
    r = upper_bound((arrvalb).begin(), (arrvalb).end(), -vala) -
        arrvalb.begin();
    l = lower_bound((arrvalb).begin(), (arrvalb).end(), -vala) -
        arrvalb.begin();
    if (arrvalb[l] != -vala) l--;
    cout << (min(abs(vala + arrvalb[l]), abs(vala + arrvalb[r]))) << endl;
  }
  return 0;
}
int main() {
  long long JUDGE_ONLINE = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Solves();
  if (!JUDGE_ONLINE)
    cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}
