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:16777216")
using namespace std;
const int inf = (int)1e9 + 5;
const long long MAXN = 300005;
const long long MAXVAL = 10000000;
const long long INF = (long long)1e17 + 50;
const long long MOD = (long long)1e9 + 7;
const unsigned long long HASH_MOD = 2305843009213693951;
string failMessage;
void fail(string addedMessage = "") {
  cout << failMessage;
  if (addedMessage != "") {
    (void)0;
    (void)0;
  }
  exit(0);
};
long long shift = 0;
set<long long> f;
long long calc_ans() {
  long long res;
  if (*f.begin() >= shift)
    res = *f.begin();
  else if (*f.rbegin() < shift)
    res = *f.rbegin();
  else {
    auto it = f.lower_bound(shift);
    res = *it - shift;
    res = min(res, abs(*(--it) - shift));
    return res;
  }
  return abs(res - shift);
}
void solve() {
  long long n, m, q;
  cin >> n >> m >> q;
  vector<long long> a(n), b(m);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long k = m - n + 1;
  long long sum_a = 0;
  for (long long i = 0; i < n; i++) {
    if (i % 2 == 0)
      sum_a += a[i];
    else
      sum_a -= a[i];
  }
  vector<long long> prefix_b(m);
  prefix_b[0] = -b[0];
  for (long long i = 1; i < m; i++) {
    if (i % 2 == 0)
      prefix_b[i] = prefix_b[i - 1] - b[i];
    else
      prefix_b[i] = prefix_b[i - 1] + b[i];
  }
  long long sum_b;
  for (long long j = 0; j < k; j++) {
    sum_b = prefix_b[j + n - 1];
    if (j > 0) sum_b -= prefix_b[j - 1];
    if (j % 2 == 1) sum_b = -sum_b;
    f.insert(sum_a + sum_b);
  }
  cout << calc_ans() << endl;
  long long x, y, z;
  for (long long it = 0; it < q; it++) {
    cin >> x >> y >> z;
    x--;
    y--;
    if ((y - x) % 2 == 0) {
      if (x % 2 == 0)
        shift -= z;
      else
        shift += z;
    }
    cout << calc_ans() << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(15);
  solve();
  (void)0;
}
