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 long long N = 100001, h = 25;
long long t[2 * N], n, m, d[N], t2[2 * N], d2[N];
long long add(long long a, long long b) {
  long long ans = a + b;
  return ans;
}
void in_build() {
  for (long long i = n - 1; i > 0; i--) {
    t[i] = (t[i << 1] + t[i << 1 | 1]);
  }
}
void apply(long long p, long long value) {
  t[p] += value;
  if (p < n) d[p] += value;
}
void build(long long p) {
  while (p > 1) p >>= 1, t[p] = add(t[p << 1], t[p << 1 | 1]) + d[p];
}
void push(long long p) {
  for (long long s = h; s > 0; --s) {
    long long i = p >> s;
    if (d[i] != 0) {
      apply(i << 1, d[i]);
      apply(i << 1 | 1, d[i]);
      d[i] = 0;
    }
  }
}
void inc(long long l, long long r, long long value) {
  l += n, r += n;
  long long l0 = l, r0 = r;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) apply(l++, value);
    if (r & 1) apply(--r, value);
  }
  build(l0);
  build(r0 - 1);
}
long long query(long long l, long long r) {
  l += n, r += n;
  push(l);
  push(r - 1);
  long long res = 0;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = add(res, t[l++]);
    if (r & 1) res = add(t[--r], res);
  }
  return res;
}
void in_build2() {
  for (long long i = n - 1; i > 0; i--) t2[i] = add(t2[i << 1], t2[i << 1 | 1]);
}
void apply2(long long p, long long value) {
  t2[p] += value;
  if (p < n) d2[p] += value;
}
void build2(long long p) {
  while (p > 1) p >>= 1, t2[p] = add(t2[p << 1], t2[p << 1 | 1]) + d2[p];
}
void push2(long long p) {
  for (long long s = h; s > 0; --s) {
    long long i = p >> s;
    if (d2[i] != 0) {
      apply2(i << 1, d2[i]);
      apply2(i << 1 | 1, d2[i]);
      d2[i] = 0;
    }
  }
}
void inc2(long long l, long long r, long long value) {
  l += n, r += n;
  long long l0 = l, r0 = r;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) apply2(l++, value);
    if (r & 1) apply2(--r, value);
  }
  build2(l0);
  build2(r0 - 1);
}
long long query2(long long l, long long r) {
  l += n, r += n;
  push2(l);
  push2(r - 1);
  long long res = 0;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = add(res, t2[l++]);
    if (r & 1) res = add(t2[--r], res);
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q, gg = 0;
  cin >> n >> m >> q;
  memset(t, 0, sizeof(t));
  memset(d, 0, sizeof(d));
  memset(d2, 0, sizeof(d2));
  memset(t2, 0, sizeof(t2));
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    if (i % 2 == 0) {
      t[n + gg] = x;
    } else {
      t2[n + gg] = x;
      gg++;
    }
  }
  long long b[m], tmp = 0, k = 0;
  for (long long i = 0; i < m; i++) cin >> b[i];
  for (long long i = 0; i < n; i++) {
    if (i % 2)
      tmp += b[i];
    else
      tmp -= b[i];
  }
  if (n % 2) k = 1;
  vector<long long> v;
  v.push_back(tmp);
  for (long long i = n; i < m; i++) {
    tmp += b[i - n];
    tmp *= -1;
    if (k % 2 == 0)
      tmp += b[i];
    else
      tmp -= b[i];
    v.push_back(tmp);
  }
  sort(v.begin(), v.end());
  in_build();
  in_build2();
  long long zz = query(0, n) - query2(0, n);
  long long lol = 1e18;
  for (auto it : v) {
    lol = min(lol, abs(zz + it));
  }
  cout << lol << "\n";
  while (q--) {
    long long l, r, val;
    cin >> l >> r >> val;
    if ((r - l + 1) % 2) {
      if (l % 2)
        zz += val;
      else
        zz -= val;
    }
    long long ans = zz, f = 1e18;
    long long l0 = 0, r0 = v.size() - 1;
    while (l0 < r0) {
      long long mid = (l0 + r0) / 2;
      tmp = ans + v[mid];
      f = min(f, abs(tmp));
      if ((v[mid] + ans) < 0)
        l0 = mid + 1;
      else
        r0 = mid;
    }
    f = min(f, abs(ans + v[l0]));
    cout << f << "\n";
  }
  return 0;
}
