//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn = 200000 * 2;
int n, m;
int step[maxn];
int l[maxn], r[maxn], belong[maxn], block, num;
int to[maxn], need[maxn];

void build(){
    block = sqrt(n); num = n / block;
    if (n % block) num++;
    for (int i = 1; i <= num; i++){
        l[i] = (i - 1) * block + 1, r[i] = i * block;
    }
    for (int i = 1; i <= n; i++)
        belong[i] = (i - 1) / block + 1;
}


void update(int be){
    for (int i = l[be]; i <= r[be]; i++)
        to[i] = -1, need[i] = 0;
    for (int i = r[be]; i >= l[be]; i--){
        int nxtpos = i + step[i];
        if (nxtpos > n) nxtpos = n + 1;
        if (belong[nxtpos] != be){//如果只走一步，且两个不是同一个块
            to[i] = nxtpos; need[i] = 1;
        }
        else {//同一个块
            need[i] = 1;
            need[i] += need[nxtpos];
            to[i] = to[nxtpos];
        }
    }
}

void query(int pos){
    int cnt = 0;
    while (pos <= n){
        cnt += need[pos];
        pos = to[pos];
    }
    printf("%d\n", cnt);
}

int main(){
    cin >> n;
    for (int i = 1; i <= n; i++){
        scanf("%d", step + i);
    }
    build();
    memset(to, -1, sizeof(to));
    for (int i = 1; i <= num; i++){
        update(i);
    }
/*
    for (int i = 1; i <= n; i++){
        printf("to[%d] = %d need[%d] = %d\n", i, to[i], i, need[i]);
    }
*/
    cin >> m;
    for (int i = 1; i <= m; i++){
        int a, b, c;
        scanf("%d%d", &a, &b); b++;
        if (a == 2){
            scanf("%d", &c);
            step[b] = c;
            update(belong[b]);
        }
        else {
            query(b);
        }
    }
    return 0;
}