#include<bits/stdc++.h>
using namespace std;
int head[100005],nxt[30000005],pos[30000005],add[30000005],id[30000005];
#define int long long
int p[2005],ad[2005],nw =0;
int pre[100005],ppre[100005],ans[100005];
int sq;int kpre[100005],npre[100005],be[100005],L[1005],R[1005],val[100005];
int tt =0 ;
int n,m;int l[100005],r[100005],a[100005];
int ll[100005],rr[100005];
void update(int rr){
	val[rr]++;
	for(int j = be[rr];j>=1;j--)kpre[j]++;
	npre[R[be[rr]]] = val[R[be[rr]]];
	for(int j = R[be[rr]]-1;j>=L[be[rr]];j--)npre[j] = val[j]+npre[j+1];
//	for(int j = 1;j<=sq;j++)cout << kpre[j] << " ";cout << endl;
//	for(int j = 1;j<=n;j++)cout << npre[j] << " ";cout << endl;
}
void init(){
	sq = sqrt(n);
	for(int i = 1;i<=sq;i++){
		L[i] = (i-1)*sq+1,R[i] = i*sq;
	}R[sq] = n;
	for(int i = 1;i<=sq;i++){
		for(int j = L[i];j<=R[i];j++)be[j] = i;
	}
}void clear_(){
	nw =0 ;
	for(int i = 1;i<=n;i++)pre[i] = pre[i-1]+a[i];
	for(int i = 1;i<=n;i++)ppre[i] = ppre[i-1]+pre[r[i]]-pre[l[i]-1];
}
signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>> n;
	for(int i = 1;i<=n;i++)cin >> a[i];
	for(int i = 1;i<=n;i++)cin >> l[i] >> r[i]; 
	init();clear_();
	cin >> m;
	for(int i = 1;i<=m;i++){
		int op;cin>> op;
		if(op == 1){
			int x,y;cin >> x >> y;
			p[++nw] = x,ad[nw] = y-a[x];a[x] = y;
			if(nw == 102)clear_();
		}else{
			int l,r;cin >> l >> r;ll[i] = l,rr[i] = r;ans[i]=ppre[r]-ppre[l-1];
			for(int j = 1;j<=nw;j++){
				id[++tt] = i;add[tt] = ad[j];pos[tt] = p[j];
			}
		}
	}memset(head,-1,sizeof(head));
	for(int i = 1;i<=tt;i++){
		int ttt = rr[id[i]];
		nxt[i] = head[ttt];head[ttt] = i;
	//	cout << " " << ttt << " " << head[ttt] << " " << add[head[ttt]] << endl;
	}/*
	相当于一个二维前缀和，要做四遍 
	*/
	for(int i = 0;i<=n;i++)val[i] = kpre[i] = npre[i] = 0;
	for(int i = 1;i<=n;i++){
		int rr = r[i];
	//	cout << " " << rr << endl;
		update(rr);
		for(int x = head[i];x!=-1;x = nxt[x]){
			ans[id[x]]+=1ll*(kpre[be[pos[x]]+1]+npre[pos[x]])*add[x];
	//		cout<< "  " << id[x] << " " << pos[x] << " " <<  (kpre[be[pos[x]]+1]+npre[pos[x]])*add[x] <<" " << add[x]<< endl;
		}
	}for(int i =0;i<=n;i++)val[i] = kpre[i] = npre[i] =0 ;
	for(int i = 1;i<=n;i++){
		int rr = l[i]-1;
		if(rr!=0){
			update(rr);
		}for(int x = head[i];x!=-1;x = nxt[x]){
			ans[id[x]]-=1ll*(kpre[be[pos[x]]+1]+npre[pos[x]])*add[x];
	//		cout<< "  " << id[x] << " " << pos[x] << " " <<  (kpre[be[pos[x]]+1]+npre[pos[x]])*add[x] <<" " << add[x]<< endl;
		}
	}memset(head,-1,sizeof(head));
	for(int i = 1;i<=tt;i++){
		int ttt = ll[id[i]]-1;
		nxt[i] = head[ttt];head[ttt] = i;
	}for(int i =0;i<=n;i++)val[i] = kpre[i] = npre[i] =0 ;
	for(int i = 1;i<=n;i++){
		int rr = r[i];
		update(rr);
		for(int x = head[i];x!=-1;x = nxt[x]){
			ans[id[x]]-=1ll*(kpre[be[pos[x]]+1]+npre[pos[x]])*add[x];
	//		cout<< "  " << id[x] << " " << pos[x] << " " <<  (kpre[be[pos[x]]+1]+npre[pos[x]])*add[x] <<" " << add[x]<< endl;
		}
	}for(int i =0;i<=n;i++)val[i] = kpre[i] = npre[i] =0 ;
	for(int i = 1;i<=n;i++){
		int rr = l[i]-1;
		if(rr!=0){
			update(rr);
		}for(int x = head[i];x!=-1;x = nxt[x]){
			ans[id[x]]+=1ll*(kpre[be[pos[x]]+1]+npre[pos[x]])*add[x];
	//		cout<< "  " << id[x] << " " << pos[x] << " " <<  (kpre[be[pos[x]]+1]+npre[pos[x]])*add[x] <<" " << add[x]<< endl;
		}
	}for(int i = 1;i<=m;i++)if(rr[i]!=0)cout << ans[i] << '\n'; 
	return 0;
}/*
nsqrtlog做法：定期重构
nsqrt只需要离线下来分块平衡一下。
链式前向星似乎有莫名其妙的大常数（在noi.ac上）
不懂啊。 
我感觉每次模拟赛我都可以想出非正解的大常数大代码难度的愚蠢做法。 
10
2067261 384717275 17463453 888985702 138961334 1411632 688969676 74515292 188541827 77102447
2 5
3 9
5 10
8 8
1 2
 3 10
 4 4
 2 3
1 5
2 4
5
1 8 604985272
1 1 800602979
1 7 927577929
2 2 7
2 6 8
*/