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>
#pragma GCC optimize "O3"
using namespace std;
constexpr int N = 100001;
int n, m, q;
long long a[N], b[N], sa, sbs[N];
long long query() {
  int p0 = lower_bound(sbs, sbs + m - n + 1, -sa) - sbs;
  if (p0 == 0) {
    return abs(sa + sbs[p0]);
  } else if (p0 == m - n + 1) {
    return abs(sa + sbs[p0 - 1]);
  } else {
    return min(abs(sa + sbs[p0]), abs(sa + sbs[p0 - 1]));
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n; ++i) scanf("%lld", a + i);
  for (int i = 0; i < n; ++i) sa += ((i & 1) ? -1 : 1) * a[i];
  for (int i = 0; i < m; ++i) scanf("%lld", b + i);
  long long sb = 0;
  for (int i = 0; i < n; ++i) sb += ((i & 1) ? 1 : -1) * b[i];
  for (int i = 0; i <= m - n; ++i) {
    sbs[i] = sb;
    sb += b[i];
    sb *= -1;
    sb += ((n & 1) ? -1 : 1) * b[i + n];
  }
  sort(sbs, sbs + m - n + 1);
  printf("%lld\n", query());
  for (int i = 0; i < q; ++i) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    --l;
    if ((l - r) & 1) sa += ((l & 1) ? -1 : 1) * x;
    printf("%lld\n", query());
  }
}
