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>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long fh = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= fh;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
long long n, m, q;
long long a[100005];
long long b[100005];
vector<long long> vec;
signed main() {
  read(n);
  read(m);
  read(q);
  for (long long i = 1; i <= n; ++i) read(a[i]);
  for (long long i = 1; i <= m; ++i) read(b[i]);
  long long cur = 0;
  for (long long i = 1; i <= n; ++i)
    if (i & 1)
      cur += b[i];
    else
      cur -= b[i];
  for (long long i = 1; i <= m - n + 1; ++i) {
    vec.push_back(cur);
    cur -= b[i];
    cur = -cur;
    if (n & 1)
      cur += b[i + n];
    else
      cur -= b[i + n];
  }
  sort(vec.begin(), vec.end());
  cur = 0;
  for (long long i = 1; i <= n; ++i)
    if (i & 1)
      cur += a[i];
    else
      cur -= a[i];
  long long id = lower_bound(vec.begin(), vec.end(), cur) - vec.begin();
  if (id == vec.size())
    writeln(cur - vec[id - 1]);
  else if (id == 0)
    writeln(vec[id] - cur);
  else
    writeln(min(vec[id] - cur, cur - vec[id - 1]));
  while (q--) {
    long long l, r, x;
    read(l);
    read(r);
    read(x);
    if ((r - l + 1) & 1) {
      if (l & 1)
        cur += x;
      else
        cur -= x;
    }
    id = lower_bound(vec.begin(), vec.end(), cur) - vec.begin();
    if (id == vec.size())
      writeln(cur - vec[id - 1]);
    else if (id == 0)
      writeln(vec[id] - cur);
    else
      writeln(min(vec[id] - cur, cur - vec[id - 1]));
  }
  return 0;
}
