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;
template <typename T>
void inpos(T &x) {
  x = 0;
  register T c = getchar();
  while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
  bool neg = false;
  if (c == '-') neg = true;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x << 3) + (x << 1) + (c & 15);
  if (neg) x = -x;
}
template <typename T>
void outpos(T n) {
  if (n < 0) {
    putchar('-');
    n *= -1;
  }
  char snum[65];
  int i = 0;
  do {
    snum[i++] = n % 10 + '0';
    n /= 10;
  } while (n);
  i = i - 1;
  while (i >= 0) putchar(snum[i--]);
  putchar('\n');
}
const int MAX = 1e5 + 5;
const int LIM = 3e5 + 5;
int n, m;
long long asum = 0;
long long be[MAX], bo[MAX];
vector<long long> sums;
void solve() {
  int pos = lower_bound(sums.begin(), sums.end(), asum) - sums.begin();
  int u = max(0, pos - 3);
  int v = min((int)sums.size() - 1, pos + 3);
  long long ans = LLONG_MAX;
  for (int i = u; i <= v; ++i) {
    long long val = asum - sums[i];
    if (val < 0) {
      val *= -1;
    }
    ans = min(ans, val);
  }
  outpos(ans);
}
int main() {
  int q;
  long long x;
  inpos(n), inpos(m), inpos(q);
  asum = 0;
  for (int i = 1; i <= n; ++i) {
    inpos(x);
    if (i % 2 == 0) {
      asum -= x;
    } else {
      asum += x;
    }
  }
  for (int i = 1; i <= m; ++i) {
    inpos(x);
    if (i % 2 == 0) {
      be[i / 2] = x;
    } else {
      bo[i / 2 + 1] = x;
    }
  }
  for (int i = 1; i <= m; ++i) {
    bo[i] += bo[i - 1];
    be[i] += be[i - 1];
  }
  long long val = 0;
  for (int i = 1; i <= m - n + 1; ++i) {
    if (n % 2 == 0) {
      if (i % 2 == 0) {
        val = be[(i + n - 1) / 2] - be[(i - 1) / 2];
        val += -bo[(i + n) / 2] + bo[i / 2];
      } else {
        val = bo[(i + n) / 2] - bo[i / 2];
        val += -be[(i + n) / 2] + be[i / 2];
      }
    } else {
      if (i % 2 == 0) {
        val = be[(i + n) / 2] - be[(i - 1) / 2];
        val += -bo[(i + n) / 2] + bo[i / 2];
      } else {
        val = bo[(i + n) / 2] - bo[i / 2];
        val += -be[(i + n - 1) / 2] + be[i / 2];
      }
    }
    sums.push_back(val);
  }
  sort(sums.begin(), sums.end());
  solve();
  int l, r;
  while (q--) {
    inpos(l), inpos(r), inpos(x);
    if ((r - l + 1) % 2 == 1) {
      if (l % 2 == 0) {
        asum -= x;
      } else {
        asum += x;
      }
    }
    solve();
  }
  return 0;
}
