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 long long maxn = 200200;
long long n, m, q;
long long a[maxn];
long long b[maxn];
long long A;
long long B[maxn];
void answer() {
  long long ans = 1e18;
  long long i = lower_bound(B, B + m - n + 1, A) - B;
  if (i < m - n + 1) ans = min(ans, B[i] - A);
  if (i) ans = min(ans, A - B[i - 1]);
  cout << ans << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  for (long long i = n; i--;) A = a[i] - A;
  for (long long i = m; i--;) B[i] = b[i] - B[i + 1];
  for (long long i = 0; i < m; i++)
    if (n % 2 == 0)
      B[i] -= B[i + n];
    else
      B[i] += B[i + n];
  sort(B, B + m - n + 1);
  answer();
  for (long long i = 0; i < q; i++) {
    long long l, r, ll, rr, x;
    cin >> l >> r >> x;
    l--, r--;
    ll = l, rr = r;
    while (ll % 2) ll++;
    while (rr % 2) rr--;
    A += x * max((rr - ll + 2) / 2, 0ll);
    ll = l, rr = r;
    while (ll % 2 == 0) ll++;
    while (rr % 2 == 0) rr--;
    A -= x * max((rr - ll + 2) / 2, 0ll);
    answer();
  }
  return 0;
}
