#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000;
int n;
long long int d[MAXN], st[2 * MAXN];
long long int arr(int i) {
  if (i == 0 || i == n - 1) return 0;
  return max(0ll, -d[i - 1]) + max(0ll, d[i]);
}
void build() {
  for (int i = 0; i < n; ++i) st[i + n] = arr(i);
  for (int i = n - 1; i > 0; --i) st[i] = min(st[2 * i], st[2 * i + 1]);
}
long long int query(int l, int r) {
  long long int res = LLONG_MAX;
  for (l += n, r += n; l < r; l /= 2, r /= 2) {
    if (l & 1) res = min(res, st[l++]);
    if (r & 1) res = min(res, st[--r]);
  }
  return res;
}
void update(int idx) {
  st[idx + n] = arr(idx);
  for (int i = idx + n; i > 1; i /= 2) st[i / 2] = min(st[i], st[i ^ 1]);
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long int sum = 0;
  int a1, a2;
  cin >> a1;
  for (int i = 0; i < n - 1; ++i)
    cin >> a2, d[i] = a2 - a1, sum += abs(d[i]), a1 = a2;
  int q;
  cin >> q;
  build();
  while (q--) {
    int t, l, r;
    long long int x;
    cin >> t >> l >> r >> x;
    l--, r--;
    if (t == 1) {
      long long int res = sum + (l == r ? abs(d[l - 1] + x) - abs(d[l - 1]) +
                                              abs(d[l] - x) - abs(d[l])
                                        : max(0ll, 2 * (x - query(l, r + 1))));
      cout << res << endl;
    } else {
      for (int i : {l - 1, r})
        sum -= abs(d[i]), d[i] += (i == l - 1 ? x : -x), sum += abs(d[i]);
      for (int i : {l - 1, l, r, r + 1}) update(i);
    }
  }
}

### Prompt

Create a solution in cpp for the following problem:

Fafa has an array A of n positive integers, the function f(A) is defined as <image>. He wants to do q queries of two types:

  * 1 l r x — find the maximum possible value of f(A), if x is to be added to one element in the range [l, r]. You can choose to which element to add x. 
  * 2 l r x — increase all the elements in the range [l, r] by value x. 



Note that queries of type 1 don't affect the array elements.

Input

The first line contains one integer n (3 ≤ n ≤ 105) — the length of the array.

The second line contains n positive integers a1, a2, ..., an (0 < ai ≤ 109) — the array elements.

The third line contains an integer q (1 ≤ q ≤ 105) — the number of queries. 

Then q lines follow, line i describes the i-th query and contains four integers ti li ri xi <image>. 

It is guaranteed that at least one of the queries is of type 1.

Output

For each query of type 1, print the answer to the query.

Examples

Input

5
1 1 1 1 1
5
1 2 4 1
2 2 3 1
2 4 4 2
2 3 4 1
1 3 3 2


Output

2
8


Input

5
1 2 3 4 5
4
1 2 4 2
2 2 4 1
2 3 4 1
1 2 4 2


Output

6
10

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000;
int n;
long long int d[MAXN], st[2 * MAXN];
long long int arr(int i) {
  if (i == 0 || i == n - 1) return 0;
  return max(0ll, -d[i - 1]) + max(0ll, d[i]);
}
void build() {
  for (int i = 0; i < n; ++i) st[i + n] = arr(i);
  for (int i = n - 1; i > 0; --i) st[i] = min(st[2 * i], st[2 * i + 1]);
}
long long int query(int l, int r) {
  long long int res = LLONG_MAX;
  for (l += n, r += n; l < r; l /= 2, r /= 2) {
    if (l & 1) res = min(res, st[l++]);
    if (r & 1) res = min(res, st[--r]);
  }
  return res;
}
void update(int idx) {
  st[idx + n] = arr(idx);
  for (int i = idx + n; i > 1; i /= 2) st[i / 2] = min(st[i], st[i ^ 1]);
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long int sum = 0;
  int a1, a2;
  cin >> a1;
  for (int i = 0; i < n - 1; ++i)
    cin >> a2, d[i] = a2 - a1, sum += abs(d[i]), a1 = a2;
  int q;
  cin >> q;
  build();
  while (q--) {
    int t, l, r;
    long long int x;
    cin >> t >> l >> r >> x;
    l--, r--;
    if (t == 1) {
      long long int res = sum + (l == r ? abs(d[l - 1] + x) - abs(d[l - 1]) +
                                              abs(d[l] - x) - abs(d[l])
                                        : max(0ll, 2 * (x - query(l, r + 1))));
      cout << res << endl;
    } else {
      for (int i : {l - 1, r})
        sum -= abs(d[i]), d[i] += (i == l - 1 ? x : -x), sum += abs(d[i]);
      for (int i : {l - 1, l, r, r + 1}) update(i);
    }
  }
}

```