#include<iostream>
#include<stack>
#include<map>
#include<vector>
#include<string>
#include<cassert>
#include<unordered_map>
#include <queue>
#include <cstdint>
#include<cstring>
#include<limits.h>
#include<cmath>
#include<set>
#include<algorithm>
#include <iomanip>
#include<numeric>
#include<bitset>
using namespace std;
#define ll long long
#define f first
#define s second
#define pii pair<int,int>
#define ppii pair<int,pii>
#define vi vector<int>
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define F(n) for(int i=0;i<n;i++)
#define lb lower_bound
#define ub upper_bound
#define fastio ios::sync_with_stdio(false);cin.tie(NULL);
#pragma GCC optimize ("03,unroll-lopps")
#define int long long
using namespace std;
const int mod=1e9+7,mxn=2e5+5,inf=1e18,minf=-1e18,lg=30;
//#undef int
int n,k,m;
int val[mxn+10],put[mxn+10],pref[mxn+10];
void setIO(string name){
ios_base::sync_with_stdio(0); cin.tie(0);
freopen((name+".in").c_str(),"r",stdin);
freopen((name+".out").c_str(),"w",stdout);
}
void solve(int l,int r){
stack<int>st,st2;
int tmp=val[l-1];
val[l-1]=inf;
st.push(l-1);
st2.push(l-1);
pref[l-1]=put[l-1]=0;
for(int i=l;i<=r;i++){
pref[i]=val[i]+pref[i-1];
put[i]=0;
while(val[i]>val[st.top()])st.pop();
if(pref[i-1]-pref[st.top()]<val[i])put[st.top()+1]++,put[i]--;
while(val[i]>=val[st2.top()]){
if(pref[i-1]-pref[st2.top()]<val[st2.top()])put[st2.top()+1]++,put[i]--;
st2.pop();
}
st.push(i);
st2.push(i);
}
while(!st2.empty()){
if(st2.top()!=l-1&&pref[r]-pref[st2.top()]<val[st2.top()]){
put[st2.top()+1]++;
}
st2.pop();
}
val[l-1]=tmp;
int ans=0;
for(int i=l;i<=r;i++){
put[i]+=put[i-1],ans+=(!put[i]);
}
cout<<ans<<'\n';
}
int32_t main(){
fastio
//brute force check
cin>>n;
for(int i=1;i<=n;i++)cin>>val[i];
int q;cin>>q;
while(q--){
int t,x,y;cin>>t>>x>>y;
if(t==1)val[x]=y;
else solve(x,y);
}
}
/*
each element will be bounded by range l,r
where sum(l+1,r-1)<minval(l,r)
then every element in range l,r cant go out of range l,r
range where x is one of the bound can be found
by finding the first left and right pos y where val[y]>=val[x]
because if val[a]>=x and a<y
then range (a,y) will also cover if range (a,l) cover
we can update range and find y for the updated x by bin search+seg tree
how do we answer?
let ql,qr be the queried range
we have list of bounded range(l,r)
if l,r is in ql,qr or l,r intersect with ql,qr then we can still use that l,r
but if l,r cover ql,qr then we can use that l,r?
*/
