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 INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 1e5 + 500;
long long int a[N], b[N], v[N];
set<long long int> st;
set<long long int>::iterator it;
long long int md(long long int a) {
  if (a < 0) return -a;
  return a;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  long long int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  v[0] = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0)
      v[0] -= b[i];
    else
      v[0] += b[i];
  }
  st.insert(v[0]);
  for (int i = n; i < m; i++) {
    v[i - n + 1] = -b[i - n] - v[i - n];
    if (n % 2 == 0)
      v[i - n + 1] += b[i];
    else
      v[i - n + 1] -= b[i];
    st.insert(v[i - n + 1]);
  }
  for (int i = 0; i < m - n; i++) st.insert(v[i]);
  long long int ac = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0)
      ac += a[i];
    else
      ac -= a[i];
  }
  long long int l = 1, r = 3, x = 0, ans;
  while (true) {
    ans = INFL;
    l--;
    r--;
    if ((r - l) % 2 == 0) {
      if (r % 2 == 0)
        ac += x;
      else
        ac -= x;
    }
    it = st.lower_bound(0);
    if (it != st.end()) {
      if (md(ans + ac) > md(*it + ac)) ans = *it;
    }
    if (it != st.begin()) it--;
    if (it != st.end()) {
      if (md(ans + ac) > md(*it + ac)) ans = *it;
    }
    it = st.lower_bound(-ac);
    if (it != st.end()) {
      if (md(ans + ac) > md(*it + ac)) ans = *it;
    }
    if (it != st.begin()) it--;
    if (it != st.end()) {
      if (md(ans + ac) > md(*it + ac)) ans = *it;
    }
    cout << md(ans + ac) << endl;
    if (q == 0) break;
    cin >> l >> r >> x;
    q--;
  }
  return 0;
}
