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 int inf = ~0u >> 1, MOD = 1e9 + 7;
inline int rd() {
  int first = 0, c = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar()) f = c != '-';
  for (; c >= '0' && c <= '9'; c = getchar()) first = first * 10 + c - '0';
  return f ? first : -first;
}
const int N = 1e5 + 11;
int n, m, q;
int a[N], b[N];
long long s[N], t[N], c[N];
int sz;
long long query(long long now) {
  long long *f = lower_bound(c + 1, c + sz + 1, now);
  if (f == c + sz + 1)
    f--;
  else if (f == c)
    f++;
  return f == c + 1 ? abs(*f - now) : min(abs(*f - now), abs(*(f - 1) - now));
}
int main() {
  n = rd(), m = rd(), q = rd();
  for (int i = (1), i_end = (n); i <= i_end; i++) {
    a[i] = rd();
    s[i] = s[i - 1] + ((i) % 2 ? a[i] : -a[i]);
  }
  for (int i = (1), i_end = (m); i <= i_end; i++) {
    b[i] = rd();
    t[i] = t[i - 1] + ((i) % 2 ? b[i] : -b[i]);
  }
  sz = m - n + 1;
  long long now = LONG_LONG_MAX;
  for (int i = (0), i_end = (m - n); i <= i_end; i++) {
    c[i + 1] = s[n] + (i % 2 ? t[i + n] - t[i] : t[i] - t[i + n]);
    now = min(now, abs(c[i + 1]));
  }
  cout << now << endl;
  sort(c + 1, c + sz + 1);
  long long tot = 0;
  while (q--) {
    int l = rd(), r = rd(), first = rd();
    if (r - l & 1)
      printf("%lld\n", now);
    else {
      if (l & 1)
        tot += first;
      else
        tot -= first;
      printf("%lld\n", now = query(-tot));
    }
  }
}
