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;
char ch, B[1 << 20], *S = B, *T = B;
long long aa, bb;
long long rd() {
  while (ch = (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T)
                   ? 0
                   : *S++),
         !(ch >= '0' && ch <= '9') && ch != '-')
    ;
  ch == '-' ? aa = bb = 0 : (aa = ch - '0', bb = 1);
  while (ch = (S == T && (T = (S = B) + fread(B, 1, 1 << 20, stdin), S == T)
                   ? 0
                   : *S++),
         (ch >= '0' && ch <= '9'))
    aa = aa * 10 + ch - '0';
  return bb ? aa : -aa;
}
const long long MAXN = 1e5 + 10;
long long n, m, q, a[MAXN], b[MAXN], suma, sumb[MAXN];
long long ask(long long l, long long r, long long x, long long delta) {
  if ((r - l) % 2 == 0) {
    if (l % 2 == 1)
      suma += x;
    else
      suma -= x;
  }
  long long *pos = lower_bound(sumb, sumb + delta, -suma);
  long long ans = 1e18 + 10;
  if (pos != sumb + delta) ans = min(ans, abs(suma + *pos));
  if (pos != sumb) {
    --pos;
    ans = min(ans, abs(suma + *pos));
  }
  return ans;
}
inline void work() {
  scanf("%lld%lld%lld", &n, &m, &q);
  suma = 0;
  for (register long long i = 1, mul = 1; i <= n; ++i, mul *= -1) {
    scanf("%lld", &a[i]);
    suma += a[i] * mul;
  }
  long long delta = m - n + 1;
  sumb[0] = 0;
  for (register long long i = 1, mul = -1; i <= m; ++i, mul *= -1) {
    scanf("%lld", &b[i]);
    if (i <= n) sumb[0] += b[i] * mul;
  }
  for (register long long i = 1; i + n <= m; ++i)
    sumb[i] = -(sumb[i - 1] + b[i]) + b[i + n] * ((n & 1) ? -1 : 1);
  sort(sumb, sumb + delta);
  printf("%lld\n", ask(1, 1, 0, delta));
  for (register long long i = 1; i <= q; ++i) {
    long long l, r, x;
    scanf("%lld%lld%lld", &l, &r, &x);
    printf("%lld\n", ask(l, r, x, delta));
  }
}
signed main() {
  work();
  return 0;
}
