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;
struct PSUM {
  int n;
  vector<long long> vll;
  PSUM(int n) : n(n) { vll.assign(n, 0); }
  void build() {
    for (int i = 1; i < n; ++i) {
      vll[i] += vll[i - 1];
    }
  }
  void set(int idx, long long val) { vll[idx] = val; }
  long long query(int lo, int hi) {
    if (lo <= 0) {
      return vll[hi];
    }
    return vll[hi] - vll[lo - 1];
  }
};
struct Segment {
  int n, idx;
  vector<long long> seg, lazy;
  Segment(int n) : n(n) {
    idx = 1;
    while (idx < n) {
      idx <<= 1;
      seg.assign(n * 4, 0);
      lazy.assign(n * 4, 0);
    }
  }
  long long update(int a, int b, long long val, int node, int l, int r) {
    if (r < a || b < l) return seg[node];
    if (lazy[node]) {
      if (l != r) {
        seg[node * 2] += lazy[node] / 2;
        lazy[node * 2] += lazy[node] / 2;
        seg[node * 2 + 1] += lazy[node] / 2;
        lazy[node * 2 + 1] += lazy[node] / 2;
      }
      lazy[node] = 0;
    }
    if (a <= l && r <= b) {
      seg[node] += val * (long long)(r - l + 1);
      lazy[node] += val * (long long)(r - l + 1);
      return seg[node];
    }
    int mid = (l + r) >> 1;
    long long lSum = update(a, b, val, node * 2, l, mid);
    long long rSum = update(a, b, val, node * 2 + 1, mid + 1, r);
    return seg[node] = lSum + rSum;
  }
  long long query(int a, int b, int node, int l, int r) {
    if (r < a || b < l) return 0;
    if (lazy[node]) {
      if (l != r) {
        seg[node * 2] += lazy[node] / 2;
        lazy[node * 2] += lazy[node] / 2;
        seg[node * 2 + 1] += lazy[node] / 2;
        lazy[node * 2 + 1] += lazy[node] / 2;
      }
      lazy[node] = 0;
    }
    if (a <= l && r <= b) return seg[node];
    int mid = (l + r) >> 1;
    long long lSum = query(a, b, node * 2, l, mid);
    long long rSum = query(a, b, node * 2 + 1, mid + 1, r);
    return lSum + rSum;
  }
  void update(int lo, int hi, long long val) {
    if (lo > hi) {
      return;
    }
    update(lo, hi, val, 1, 0, idx - 1);
  }
  long long query(int lo, int hi) {
    if (lo > hi) {
      return 0;
    }
    return query(lo, hi, 1, 0, idx - 1);
  }
};
int n, m, x, y, q;
long long z;
set<long long> B;
long long ABS(long long num) { return num < 0 ? -num : num; }
int getGteEven(int num) { return num % 2 == 0 ? num : num + 1; }
int getGteOdd(int num) { return num % 2 == 1 ? num : num + 1; }
int getLteEven(int num) { return num % 2 == 0 ? num : num - 1; }
int getLteOdd(int num) { return num % 2 == 1 ? num : num - 1; }
long long solve(Segment& odd, Segment& even) {
  long long ans = 98765432123456788LL, sum = even.query(0, n) - odd.query(0, n);
  auto it = B.lower_bound(-sum);
  if (it != B.end()) {
    ans = min(ans, ABS(sum + *it));
  }
  if (it != B.begin()) {
    --it;
    ans = min(ans, ABS(sum + *it));
  }
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  Segment odd(n + 1), even(n + 1);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &z);
    if (i & 1) {
      odd.update(i / 2, i / 2, z);
    } else {
      even.update(i / 2, i / 2, z);
    }
  }
  PSUM startOdd(m), startEven(m);
  for (int i = 0; i < m; ++i) {
    scanf("%lld", &z);
    startEven.set(i, (i & 1) ? z : -z);
    startOdd.set(i, (i & 1) ? -z : z);
  }
  startEven.build();
  startOdd.build();
  for (int i = 0; i <= m - n; ++i) {
    int lo = i, hi = i + n - 1;
    if (i & 1) {
      B.insert(startOdd.query(lo, hi));
    } else {
      B.insert(startEven.query(lo, hi));
    }
  }
  printf("%lld\n", solve(odd, even));
  while (q--) {
    scanf("%d%d%lld", &x, &y, &z);
    --x;
    --y;
    if (getGteOdd(x) <= getLteOdd(y)) {
      odd.update(getGteOdd(x) / 2, getLteOdd(y) / 2, z);
    }
    if (getGteEven(x) <= getLteEven(y)) {
      even.update(getGteEven(x) / 2, getLteEven(y) / 2, z);
    }
    printf("%lld\n", solve(odd, even));
  }
}
