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 N = 1e5 + 7;
const int MX = 1e9 + 7;
const int INF = 1e9 + 9;
int n, m, q;
long long int a[N], b[N];
long long int suf[N];
long long int curAdd;
set<long long int> S;
long long int answer() {
  auto x = S.lower_bound(-curAdd);
  long long int ret = 1LL * INF * INF;
  if (x != S.end()) ret = min(ret, abs((*x) + curAdd));
  if (x != S.begin()) {
    --x;
    ret = min(ret, abs((*x) + curAdd));
  }
  return ret;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  long long int s = 0LL;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    s += (i & 1) ? a[i] : -a[i];
  }
  for (int i = 1; i <= m; ++i) scanf("%lld", &b[i]);
  for (int i = m; i >= 1; --i) suf[i] = b[i] - suf[i + 1];
  for (int j = 0; j <= m - n; ++j)
    S.insert(s - suf[j + 1] - suf[j + n + 1] * ((n & 1) ? 1 : -1));
  printf("%lld\n", answer());
  while (q--) {
    long long int l, r, x;
    scanf("%lld %lld %lld", &l, &r, &x);
    l %= 2;
    r %= 2;
    curAdd += (l + r - 1) * x;
    printf("%lld\n", answer());
  }
  return 0;
}
