//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 =  * ;

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 = ; i <= num; i++){

        l[i] = (i - ) * block + , r[i] = i * block;

    }

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

        belong[i] = (i - ) / block + ;

}

void update(int be){

    for (int i = l[be]; i <= r[be]; i++)

        to[i] = -, need[i] = ;

    for (int i = r[be]; i >= l[be]; i--){

        int nxtpos = i + step[i];

        if (nxtpos > n) nxtpos = n + ;

        if (belong[nxtpos] != be){//如果只走一步，且两个不是同一个块

            to[i] = nxtpos; need[i] = ;

        }

        else {//同一个块

            need[i] = ;

            need[i] += need[nxtpos];

            to[i] = to[nxtpos];

        }

    }

}

void query(int pos){

    int cnt = ;

    while (pos <= n){

        cnt += need[pos];

        pos = to[pos];

    }

    printf("%d\n", cnt);

}

int main(){

    cin >> n;

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

        scanf("%d", step + i);

    }

    build();

    memset(to, -, sizeof(to));

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

        int a, b, c;

        scanf("%d%d", &a, &b); b++;

        if (a == ){

            scanf("%d", &c);

            step[b] = c;

            update(belong[b]);

        }

        else {

            query(b);

        }

    }

    return ;

}