Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i].

Unfortunately, the longer he learns, the fewer he gets.

That means, if he reads books from ll to rr, he will get a[l] \times L + a[l+1] \times (L-1) + \cdots + a[r-1] \times 2 + a[r]a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r] (LL is the length of [ ll, rr ] that equals to r - l + 1r−l+1).

Now Ryuji has qq questions, you should answer him:

11. If the question type is 11, you should answer how much knowledge he will get after he reads books [ ll, rr ].

22. If the question type is 22, Ryuji will change the ith book's knowledge to a new value.

Input

First line contains two integers nn and qq (nn, q \le 100000q≤100000).

The next line contains n integers represent a[i]( a[i] \le 1e9)a[i](a[i]≤1e9) .

Then in next qq line each line contains three integers aa, bb, cc, if a = 1a=1, it means question type is 11, and bb, ccrepresents [ ll , rr ]. if a = 2a=2 , it means question type is 22 , and bb, cc means Ryuji changes the bth book' knowledge to cc

Output

For each question, output one line with one integer represent the answer.

 //H-树状数组

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<bitset>

 #include<cassert>

 #include<cctype>

 #include<cmath>

 #include<cstdlib>

 #include<ctime>

 #include<deque>

 #include<iomanip>

 #include<list>

 #include<map>

 #include<queue>

 #include<set>

 #include<stack>

 #include<vector>

 using namespace std;

 typedef long long ll;

 const double PI=acos(-1.0);

 const double eps=1e-;

 const ll mod=1e9+;

 const int inf=0x3f3f3f3f;

 const int maxn=1e5+;

 const int maxm=1e3+;

 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

 ll a[maxn],b[maxn],n,q;

 int lowbit(int x)

 {

     return x&(-x);

 }

 void add(ll a[],int x,ll val)

 {

     for(int i=x;i<=n;i+=lowbit(i))

         a[i]+=val;

 }

 ll query(ll a[],int x)

 {

     ll ans=;

     for(int i=x;i>;i-=lowbit(i))

         ans+=a[i];

     return ans;

 }

 int main()

 {

     cin>>n>>q;

     for(int i=;i<=n;i++){

         ll val;

         cin>>val;

         add(a,i,val);//单纯的保存，类似前缀和

         add(b,i,i*val);//这样保存，减去的时候正好满足条件，*L，*(L-1)。。。

     }

     while(q--){

         ll op,l,r;

         cin>>op>>l>>r;

         if(op==){

             ll cnt=query(a,l)-query(a,l-);//单点更新

             add(a,l,r-cnt);

             ll ret=query(b,l)-query(b,l-);//同上

             add(b,l,l*r-ret);

         }

         else{

             ll cnt=(r+)*(query(a,r)-query(a,l-));//先算出横坐标的和，然后*(r+1)就是一个大矩形的面积

             ll ret=query(b,r)-query(b,l-);//倒着的梯形面积，(因为从1开始的，所以是梯形不是三角形)

             cout<<cnt-ret<<endl;

         }

     }

 }

ll query(int L,int R,int l,int r,int rt)

{

    if(L<=l&&r<=R){

        int tmp=Len;

        Len-=len[rt];

        return ans[rt]+sum[rt]*(tmp-len[rt]);

    }

    int m=(l+r)>>;

    ll ret=;

    if(L<=m) ret+=query(L,R,lson);

    if(R >m) ret+=query(L,R,rson);

    return ret;

}

 //H-线段树

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<bitset>

 #include<cassert>

 #include<cctype>

 #include<cmath>

 #include<cstdlib>

 #include<ctime>

 #include<deque>

 #include<iomanip>

 #include<list>

 #include<map>

 #include<queue>

 #include<set>

 #include<stack>

 #include<vector>

 using namespace std;

 typedef long long ll;

 const double PI=acos(-1.0);

 const double eps=1e-;

 const ll mod=1e9+;

 const int inf=0x3f3f3f3f;

 const int maxn=1e5+;

 const int maxm=1e3+;

 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

 #define lson l,m,rt<<1

 #define rson m+1,r,rt<<1|1

 ll sum[maxn<<],ans[maxn<<],len[maxn<<];

 //sum为单点的价值(横坐标)，ans为总价值，len为区间长度

 int Len;

 void pushup(int rt)

 {

     ans[rt]= ans[rt<<]+ans[rt<<|]+sum[rt<<]*len[rt<<|];//总价值为左儿子+右儿子+左儿子区间和*右儿子区间长

     len[rt]=len[rt<<]+len[rt<<|];

     sum[rt]=sum[rt<<]+sum[rt<<|];

 }

 void build(int l,int r,int rt)

 {

     if(l==r){

         scanf("%lld",&sum[rt]);

         len[rt]=;

         ans[rt]=sum[rt];

         return;

     }

     int m=(l+r)>>;

     build(lson);

     build(rson);

     pushup(rt);

 }

 ll query(int L,int R,int l,int r,int rt)

 {

     if(L<=l&&r<=R){

         int tmp=Len;

         Len-=len[rt];

         return ans[rt]+sum[rt]*(tmp-len[rt]);//求小三角形的面积就是大的梯形-平行四边形

     }

     int m=(l+r)>>;

     ll ret=;

     if(L<=m) ret+=query(L,R,lson);

     if(R >m) ret+=query(L,R,rson);

     return ret;

 }

 void update(int p,ll val,int l,int r,int rt)//单点更新

 {

     if(l==r){

         sum[rt]=val;

         ans[rt]=val;

         return;

     }

     int m=(l+r)>>;

     if(p<=m) update(p,val,lson);

     else     update(p,val,rson);

     pushup(rt);

 }

 int main()

 {

     int n, m;

     while(~scanf("%d%d",&n,&m)){

         build(,n,);

         while(m--){

             int op;

             scanf("%d",&op);

             if(op==){

                 int l,r;

                 scanf("%d%d",&l,&r);

                 Len=r-l+;

                 printf("%lld\n",query(l,r,,n,));

             }

             else{

                 int pos;

                 ll val;

                 scanf("%d%lld",&pos,&val);

                 update(pos,val,,n,);

             }

         }

     }

     return ;

 }