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 double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
long long n, q, sum[MAXN], m, l, r, k, x[MAXN], s[MAXN][2];
struct Node {
  int l, r;
  long long s, laz;
} node[MAXN * 4][2];
inline void up(int rt, int opt) {
  node[rt][opt].s = node[(rt << 1)][opt].s + node[(rt << 1 | 1)][opt].s;
}
inline void down(int rt, int opt) {
  if (node[rt][opt].laz) {
    node[(rt << 1)][opt].laz += node[rt][opt].laz;
    node[(rt << 1 | 1)][opt].laz += node[rt][opt].laz;
    node[(rt << 1)][opt].s += node[rt][opt].laz * (node[(rt << 1)][opt].r -
                                                   node[(rt << 1)][opt].l + 1);
    node[(rt << 1 | 1)][opt].s +=
        node[rt][opt].laz *
        (node[(rt << 1 | 1)][opt].r - node[(rt << 1 | 1)][opt].l + 1);
    node[rt][opt].laz = 0;
  }
}
void build(int rt, int l, int r, int opt) {
  node[rt][opt].l = l, node[rt][opt].r = r;
  node[rt][opt].s = node[rt][opt].laz = 0;
  int mid = ((l + r) >> 1);
  if (l == r) {
    node[rt][opt].s = s[l][opt];
    return;
  }
  build((rt << 1), l, mid, opt);
  build((rt << 1 | 1), mid + 1, r, opt);
  up(rt, opt);
}
void update(int rt, int l, int r, long long k, int opt) {
  if (node[rt][opt].l == l && node[rt][opt].r == r) {
    node[rt][opt].laz += k;
    node[rt][opt].s += (r - l + 1) * k;
    return;
  }
  down(rt, opt);
  int mid = ((node[rt][opt].l + node[rt][opt].r) >> 1);
  if (r <= mid)
    update((rt << 1), l, r, k, opt);
  else if (l > mid)
    update((rt << 1 | 1), l, r, k, opt);
  else {
    update((rt << 1), l, mid, k, opt);
    update((rt << 1 | 1), mid + 1, r, k, opt);
  }
  up(rt, opt);
}
long long query(int rt, int l, int r, int opt) {
  if (l > r) return 0;
  if (node[rt][opt].l == l && node[rt][opt].r == r) return node[rt][opt].s;
  down(rt, opt);
  int mid = ((node[rt][opt].l + node[rt][opt].r) >> 1);
  if (r <= mid)
    return query((rt << 1), l, r, opt);
  else if (l > mid)
    return query((rt << 1 | 1), l, r, opt);
  else
    return query((rt << 1), l, mid, opt) +
           query((rt << 1 | 1), mid + 1, r, opt);
}
long long cal() {
  long long ret = 0;
  if (n & 1) {
    int p1 = (n + 1) / 2;
    int p2 = p1 - 1;
    ret = query(1, 1, p1, 1) - query(1, 1, p2, 0);
  } else {
    int p1 = n / 2;
    int p2 = p1;
    ret = query(1, 1, p1, 1) - query(1, 1, p2, 0);
  }
  return ret;
}
void work(int cnt) {
  long long ans;
  long long tmp = cal();
  int p1 = lower_bound(x + 1, x + 1 + cnt, -tmp) - x;
  int p2 = p1 - 1;
  p1 = min(p1, cnt);
  p2 = max(1, p2);
  ans = min(abs(x[p1] + tmp), abs(x[p2] + tmp));
  cout << ans << endl;
}
int main() {
  long long tmp;
  cin >> n >> m >> q;
  int s0 = 0, s1 = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &tmp);
    if (i & 1)
      s[++s1][1] = tmp;
    else
      s[++s0][0] = tmp;
  }
  build(1, 1, s1, 1);
  if (s0 > 0) build(1, 1, s0, 0);
  for (int i = 1; i <= m; i++) {
    scanf("%lld", &tmp);
    if (i & 1)
      sum[i] = sum[i - 1] - tmp;
    else
      sum[i] = sum[i - 1] + tmp;
  }
  int cnt = 0;
  for (int i = 1; i <= m; i++) {
    int j = i + n - 1;
    if (j > m) break;
    if (i & 1)
      x[++cnt] = sum[j] - sum[i - 1];
    else
      x[++cnt] = -(sum[j] - sum[i - 1]);
  }
  sort(x + 1, x + 1 + cnt);
  work(cnt);
  int s2, t1, t2;
  while (q--) {
    scanf("%lld%lld%lld", &l, &r, &k);
    if (l & 1) {
      s1 = (l + 1) / 2;
      s2 = s1;
    } else {
      s1 = l / 2 + 1;
      s2 = l / 2;
    }
    if (r & 1) {
      t1 = (r + 1) / 2;
      t2 = t1 - 1;
    } else {
      t1 = r / 2;
      t2 = r / 2;
    }
    if (t1 >= s1) update(1, s1, t1, k, 1);
    if (t2 >= s2) update(1, s2, t2, k, 0);
    work(cnt);
  }
  return 0;
}
