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;
long long bsum[100100];
int b[100100];
vector<long long> bv;
long long asum;
int k;
long long solve() {
  auto y = lower_bound(bv.begin(), bv.end(), asum);
  long long ans;
  if (y == bv.begin()) {
    ans = abs(asum - bv[0]);
  } else if (y == bv.end()) {
    ans = abs(asum - bv[k]);
  } else {
    ans = min(abs(asum - *y), abs(asum - *(y - 1)));
  }
  return ans;
}
int main() {
  int n, m, q;
  scanf("%d%d%d", &(n), &(m), &(q));
  k = m - n;
  int sg = 1;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &(a));
    asum += (long long)sg * a;
    sg *= -1;
  }
  int bsg = 1;
  int j = 0;
  for (int i = 0; i < m; i++) {
    int bb;
    scanf("%d", &(bb));
    b[i] = bsg * bb;
    bsg *= -1;
    if (i >= n) {
      j++;
      bsum[j] = bsum[j - 1] - b[i - n];
    }
    bsum[j] += b[i];
  }
  long long sgsum = 1;
  for (int i = 0; i <= k; i++) {
    bv.push_back(sgsum * bsum[i]);
    sgsum *= -1;
  }
  sort(bv.begin(), bv.end());
  printf(
      "%lld"
      "\n",
      solve());
  for (int i = 0; i < q; i++) {
    int l, r, x;
    scanf("%d%d%d", &(l), &(r), &(x));
    l--;
    if ((r - l) % 2 == 1) {
      if (l % 2 == 0)
        asum += x;
      else
        asum -= x;
    }
    printf(
        "%lld"
        "\n",
        solve());
  }
  return 0;
}
