题意：
带修改求区间\(mex\)

思路：
在知道要用主席树求区间\(mex\)之后，只需要写一个动态主席树即可。具体思路在2021 ICPC 昆明 M里已经写了（令人遗憾的补题顺序）。
这里用树状数组套主席树。树状数组里表示不同树的根节点，也就是说，树状数组里的每一个点都是一棵树。每次修改或者询问都按照树状数组来写。注意空间要给够。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 2e5 + 7;

int n, m, q, li, ri;
int a[N], rt[N];

inline int lowbit(int x) {
    return x & -x;
}

struct Seg{
    struct Node{
        int ls, rs, cnt;
        ll sum;
        void init() {
            ls = rs = cnt = 0;
            sum = 0;
        }
    }t[N * 80];
    int tot;
    void init() {
        tot = 0;
        t[tot].init();
    }
    int newnode() {
        t[++tot].init();
        return tot;
    }
    void build(int &id, int l, int r) {
        if (!id) id = newnode();
        if (l == r) {
            return;
        }
        int mid = l + r >> 1;
        build(t[id].ls, l, mid);
        build(t[id].rs, mid + 1, r);
    }
    void modify(int &rt, int l, int r, int pos, ll v) {
        if (!rt) rt = newnode();
        t[rt].sum += v;
        if (l == r) {
            return;
        }
        int mid = l + r >> 1;
        if (pos <= mid) {
            modify(t[rt].ls, l, mid, pos, v);
        } else {
            modify(t[rt].rs, mid + 1, r, pos, v);
        }
    }
    void update(int x, int pos, ll v) {
        for (; x <= n; x += lowbit(x)) {
            modify(rt[x], 1, m, pos, v);
        }
    }
    ll query(int rt, int l, int r, ll k) {
        if (!rt) return 0;
        if (r <= k) {
//        	printf("%d %d %d %d\n", l, r, k, t[rt].sum);
            return t[rt].sum;
        }
        if (l > k) return 0;
        int mid = l + r >> 1;
        if (k <= mid) return query(t[rt].ls, l, mid, k);
        else return query(t[rt].ls, l, mid, k) + query(t[rt].rs, mid + 1, r, k);
    }

    ll query(int l, int r, ll k) {
        ll res = 0;
        for (int i = r; i > 0; i -= lowbit(i)) {
            res += query(rt[i], 1, m, k);
        }
        for (int i = l; i > 0; i -= lowbit(i)) {
            res -= query(rt[i], 1, m, k);
        }
        return res;
    }
}seg;

int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    m = 2e5;
    seg.init();
    rt[0] = 0;
    seg.build(rt[0], 1, m);
    for (int i = 1; i <= n; ++i) {
        seg.update(i, a[i], a[i]);
    }
    ll pre = 0;
    for (int i = 1, op; i <= q; ++i) {
        scanf("%d%d%d", &op, &li, &ri);
        if (op == 1) {
            seg.update(li, a[li], -a[li]);
            a[li] = ri;
            seg.update(li, a[li], a[li]);
        } else {
            pre = 1;
            ll tmp = seg.query(li - 1, ri, pre);
            while (tmp >= pre) {
                pre = tmp + 1;
                tmp = seg.query(li - 1, ri, pre);
            }
            printf("%lld\n", pre);
        }
    }
    return 0;
}