Fafa has an array A of n positive integers, the function f(A) is defined as ${\displaystyle \underset{i=1}{\overset{n}{\sum }}}$

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 .

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

題意：

給你n個數，對於f(a)的定義見題目，給你兩個操作，1 l r x 代表你可以在l到r之間選一個數加上x，讓f(a)最大，或者可以不加，2 l r x 代表在l到r之間的所有數都加上x。

題解：

對於第一個操作，我們可以知道，如果 $a_{i-1}-a_i$>0的時候，增加ai會減小f(a)，在 $a_i-a_{i+1}$<0的時候，增加ai會減小f(a),那麼我們只要維護當 $a_{i-1}-a_i$>0, $a_i-a_{i+1}$<0的時候的和，之後若是小於x，那麼就是ans+2*(x-sum).第二個操作，改變的只是l,l-1,r,r+1範圍的值，稍微修改一下即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e5+5;
ll a[N],de[N],sum[N<<2];
void update(int l,int r,int root,int pos)
{
    if(l==r)
    {
        ll ans=0;
        if(de[pos-1]>0)ans+=de[pos-1];
        if(de[pos]<0)ans-=de[pos];
        sum[root]=ans;
        return ;
    }
    int mid=l+r>>1;
    if(mid>=pos)
        update(l,mid,root<<1,pos);
    else
        update(mid+1,r,root<<1|1,pos);
    sum[root]=min(sum[root<<1],sum[root<<1|1]);
}
ll query(int l,int r,int root,int ql,int qr)
{
    if(l>=ql&&r<=qr)
        return sum[root];
    int mid=l+r>>1;
    ll ans=1e17;
    if(mid>=ql)
        ans=query(l,mid,root<<1,ql,qr);
    if(mid<qr)
        ans=min(ans,query(mid+1,r,root<<1|1,ql,qr));
    return ans;
}
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    ll all=0;
    for(int i=1;i<n;i++)
    {
        de[i]=a[i]-a[i+1];
        all+=abs(de[i]);
        if(i>1)
            update(1,n,1,i);
    }
    int op,l,r;
    ll x;
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d%d%d%lld",&op,&l,&r,&x);
        if(op==1)
        {
            if(l==r)
                printf("%lld\n",all-abs(de[l-1])-abs(de[l])+abs(de[l-1]-x)+abs(de[l]+x));
            else
            {
                ll ans=query(1,n,1,l,r);
                if(ans>x)
                    printf("%lld\n",all);
                else
                    printf("%lld\n",all+2*(x-ans));
            }
        }
        else
        {
            all-=(abs(de[l-1])+abs(de[r]));
            de[l-1]-=x,de[r]+=x;
            all+=(abs(de[l-1])+abs(de[r]));
            update(1,n,1,l),update(1,n,1,r);
            if(l-1>1)
                update(1,n,1,l-1);
            if(r+1<n)
                update(1,n,1,r+1);
        }
    }
    return 0;
}