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 maxn = 100001;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int _read() {
  char ch = nc();
  int sum = 0;
  bool flag = 1;
  while (ch != '-' && !(ch >= '0' && ch <= '9')) ch = nc();
  if (ch == '-') flag = 0, ch = nc();
  while (ch >= '0' && ch <= '9') sum = sum * 10 + ch - 48, ch = nc();
  return flag ? sum : -sum;
}
struct E {
  int N, M, Q, n;
  long long sum[maxn], b[maxn], suma;
  void init() {
    long long t;
    N = _read();
    M = _read();
    Q = _read();
    n = M - N;
    for (int i = 1, j = 1; i <= N; i++, j = -j) suma += _read() * j;
    for (int i = 1, j = 1; i <= M; i++, j = -j) b[i] = _read() * j;
    for (int i = 1; i <= N; i++) sum[0] += b[i];
    for (int i = 1, j = N + 1, k = 1; i <= n; i++, j++, k++)
      sum[i] = sum[i - 1] + b[j] - b[k];
    for (int i = 1; i <= n; i += 2) sum[i] = -sum[i];
    sort(sum, sum + n + 1);
  }
  long long binary_search(long long val) {
    int id = lower_bound(sum, sum + n + 1, val) - sum;
    if (id == 0) return sum[0] - val;
    if (id == n + 1) return val - sum[n];
    return min(sum[id] - val, val - sum[id - 1]);
  }
  void work() {
    long long l, r, x;
    printf("%I64d\n", binary_search(suma));
    while (Q--) {
      l = _read();
      r = _read();
      x = _read();
      l %= 2;
      r %= 2;
      if (l == 0 && r == 0) suma -= x;
      if (l == 1 && r == 1) suma += x;
      printf("%I64d\n", binary_search(suma));
    }
  }
} sol;
int main() {
  sol.init();
  sol.work();
  return 0;
}
