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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int nmax = 1000100;
long long res, cnt = 1e+16;
double eps = 1e-5;
long long n, q, k, l, r, m, sm, a, b, x, c, n1, d;
long long el[100010];
set<long long> ans;
vector<long long> vr;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  a = 1;
  for (int i = 1; i <= n; ++i) {
    cin >> b;
    sm += b * a;
    a *= -1;
  }
  for (int i = 1; i <= m; ++i) {
    cin >> el[i];
  }
  a = -1;
  for (int i = 1; i <= n; ++i) {
    l += el[i] * a;
    a *= -1;
  }
  a *= -1;
  ans.insert(l);
  for (int i = n + 1; i <= m; ++i) {
    l += el[i - n];
    l *= -1;
    l += a * el[i];
    ans.insert(l);
  }
  auto it = ans.lower_bound(-sm);
  if (it == ans.end()) {
    --it;
    d = *it;
    cout << abs(d + sm) << endl;
  } else {
    if (it != ans.begin()) {
      k = *it;
      --it;
      d = *it;
      cout << min(abs(d + sm), abs(k + sm)) << endl;
    } else {
      k = *it;
      cout << abs(k + sm) << endl;
    }
  }
  for (int i = 0; i < q; ++i) {
    cin >> l >> r >> x;
    if ((r - l + 1) & 1) {
      if (r & 1)
        sm += x;
      else
        sm -= x;
    }
    auto it = ans.lower_bound(-sm);
    if (it == ans.end()) {
      --it;
      d = *it;
      cout << abs(d + sm) << endl;
    } else {
      if (it != ans.begin()) {
        k = *it;
        --it;
        d = *it;
        cout << min(abs(d + sm), abs(k + sm)) << endl;
      } else {
        k = *it;
        cout << abs(k + sm) << endl;
      }
    }
  }
  return 0;
}
