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 = 1e5 + 5;
const int MAXM = 1e5 + 5;
const int mod = 1e9 + 7;
int n, m, q;
int x[MAXN], y[MAXN];
set<long long> s;
set<long long>::iterator it;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  long long sum1 = 0, sum2 = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
    if (i % 2 == 0)
      sum1 -= x[i];
    else
      sum1 += x[i];
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &y[i]);
    if (i > n) continue;
    if (i % 2 == 0)
      sum2 -= y[i];
    else
      sum2 += y[i];
  }
  s.insert(sum2);
  int bas = 1;
  for (int i = 1; i <= m - n; i++) {
    if (i % 2 == 0)
      sum2 += y[i];
    else
      sum2 -= y[i];
    if ((n + i) % 2 == 0)
      sum2 -= y[n + i];
    else
      sum2 += y[n + i];
    bas *= -1;
    s.insert(sum2 * bas);
  }
  for (int i = 0; i <= q; i++) {
    int a, b;
    long long c;
    if (i != 0) {
      scanf("%d%d%lld", &a, &b, &c);
      if (a % 2 == 0 && b % 2 == 0) sum1 -= c;
      if (a % 2 == 1 && b % 2 == 1) sum1 += c;
    }
    it = s.lower_bound(sum1);
    long long d = *it;
    long long ans = 1e18;
    if (it == s.end()) {
      it--;
    } else {
      it++;
      if (it != s.end()) {
        d = *it;
        ans = min(abs(sum1 - d), ans);
      }
      it--;
    }
    d = *it;
    ans = min(abs(sum1 - d), ans);
    if (it != s.begin()) {
      it--;
      d = *it;
      ans = min(abs(sum1 - d), ans);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
