#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const int N = 1e5 + 5;

int n, m;
int a[N];
ll tr[N * 4];

// 向上合并信息，父节点的mask是两个子节点的mask的并集
void psh(int u) {
    tr[u] = tr[u << 1] | tr[u << 1 | 1];
}

// 构建线段树
void bld(int u, int l, int r) {
    if (l == r) {
        tr[u] = 1LL << (a[l] - 1);
        return;
    }
    int mid = (l + r) >> 1;
    bld(u << 1, l, mid);
    bld(u << 1 | 1, mid + 1, r);
    psh(u);
}

// 单点修改
void upd(int u, int l, int r, int x, int v) {
    if (l == r) {
        tr[u] = 1LL << (v - 1);
        return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid) {
        upd(u << 1, l, mid, x, v);
    } else {
        upd(u << 1 | 1, mid + 1, r, x, v);
    }
    psh(u);
}

// 区间查询
ll ask(int u, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) {
        return tr[u];
    }
    int mid = (l + r) >> 1;
    ll res = 0;
    if (ql <= mid) {
        res |= ask(u << 1, l, mid, ql, qr);
    }
    if (qr > mid) {
        res |= ask(u << 1 | 1, mid + 1, r, ql, qr);
    }
    return res;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }

    bld(1, 1, n);

    while (m--) {
        int op;
        cin >> op;
        if (op == 1) {
            int l, r;
            cin >> l >> r;
            ll res = ask(1, 1, n, l, r);
            cout << __builtin_popcountll(res) << "\n";
        } else {
            int x, y;
            cin >> x >> y;
            upd(1, 1, n, x, y);
        }
    }

    return 0;
}