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;
template <typename T>
inline const T read1() {
  T m = 0;
  char k = getchar();
  while (('0' > k || k > '9') && (k != '-')) k = getchar();
  const bool f = (k == '-' ? 1 : 0);
  if (f) k = getchar();
  while ('0' <= k && k <= '9')
    m = (m << 3) + (m << 1) + (k ^ 48), k = getchar();
  return f ? -m : m;
}
int n, m, q, b[100001];
set<long long> w;
long long sa, sb;
long long Find(long long x) {
  set<long long>::iterator i = w.lower_bound(x);
  if (i != w.end()) {
    long long t = *i;
    if (i != w.begin()) return min(t - x, x - *--i);
    return t - x;
  }
  return x - *--i;
}
int main() {
  n = read1<int>(), m = read1<int>(), q = read1<int>();
  for (int i = 1, w = 1; i <= n; ++i, w = -w) sa += read1<int>() * w;
  for (int i = 1, w = 1; i <= m; ++i, w = -w)
    b[i] = read1<int>(), i <= n && (sb += b[i] * w);
  w.insert(sb);
  for (int i = n + 1, o = n & 1 ? 1 : -1; i <= m; ++i)
    sb = -sb, w.insert(sb += b[i - n] + b[i] * o);
  printf("%lld\n", Find(sa));
  for (int i = 1; i <= q; ++i) {
    int l = read1<int>(), r = read1<int>(), v = read1<int>();
    if ((l ^ ~r) & 1) sa += l & 1 ? v : -v;
    printf("%lld\n", Find(sa));
  }
  return 0;
}
