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>
bool debug = 0;
const double PI = acos(-1.0);
const double eps = 1e-9;
using namespace std;
long long b[100100];
long long a;
vector<long long> sum;
int n, m, q;
long long bs() {
  int i = 0;
  int f = sum.size() - 1;
  int m;
  if (debug)
    cout << "a"
         << " = " << (a) << endl;
  while (f > i) {
    m = (i + f) / 2;
    if (debug)
      cout << "i"
           << " = " << (i) << endl;
    if (debug)
      cout << "m"
           << " = " << (m) << endl;
    if (debug)
      cout << "f"
           << " = " << (f) << endl;
    if (debug)
      cout << "a + sum[m]"
           << " = " << (a + sum[m]) << endl;
    if (a + sum[m] >= 0) {
      f = m;
    } else
      i = m + 1;
  }
  long long res = abs(a + sum[0]);
  for (int j = -1; j <= 1; j++) {
    if (j + f < 0 || j + f > sum.size() - 1) continue;
    if (abs(a + sum[j + f]) < res) res = abs(a + sum[j + f]);
  }
  return res;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  long long k = 0;
  long long x;
  a = 0;
  k = 1LL;
  for (int i = 0; i < (n); i++) {
    scanf("%lld", &x);
    a += k * x;
    k *= -1LL;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &b[i]);
  }
  k = -1LL;
  x = 0;
  for (int i = 1; i <= n; i++) {
    x += b[i] * k;
    k *= -1LL;
  }
  sum.push_back(x);
  for (int j = 1; j <= m - n; j++) {
    x += b[1 + j - 1];
    x *= -1LL;
    if (n % 2 == 0) {
      x += b[n + j];
    } else
      x -= b[n + j];
    sum.push_back(x);
  }
  sort(sum.begin(), sum.end());
  sum.resize(distance(sum.begin(), unique(sum.begin(), sum.end())));
  for (long long i : sum) {
    if (debug)
      cout << "i"
           << " = " << (i) << endl;
  }
  printf("%lld\n", bs());
  int k1, k2;
  for (int i = 0; i < q; i++) {
    scanf("%d %d %lld", &k1, &k2, &x);
    if ((k2 - k1 + 1) % 2 == 1) {
      if (k1 % 2 == 0) {
        a -= x;
      } else
        a += x;
    }
    printf("%lld\n", bs());
  }
  return 0;
}
