带修莫队，复杂度应该是不对的，但还是厚颜无耻的来请人帮忙卡常。

感谢火车司机的快读板子。

#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
namespace IO {
    char buf[1 << 22], Out[1 << 22], *p1 = buf, *p2 = buf;
    int p3 = -1, f = 0;
    inline char getchar(){return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1 ++;}
    inline void flush(){fwrite(Out, 1, p3 + 1, stdout);p3 = -1;}
    inline void check(){if(p3 > (1 << 21)) flush();}
    inline void putchar(char a){Out[++ p3] = a;check();}
}
template <typename T>
inline void read(T &x){
    x = 0;int fu = 1;
    char c = IO :: getchar();
    while(c > 57 || c < 48){
        if(c == 45) fu = -1;
        c = IO :: getchar();
    }
    while(c <= 57 && c >= 48){
        x = (x << 3) + (x << 1) + c - 48;
        c = IO :: getchar();
    }
    x *= fu;
}
template <typename T>
inline void fprint(T x){
    if(x < 0) IO :: putchar(45), x = -x;
    if(x > 9) fprint(x / 10);
    IO :: putchar(x % 10 + 48);
}
template <typename T>
inline void fprint(T x, char ch){
    fprint(x);IO :: putchar(ch);
}
inline char next_char(){
    char ch = IO :: getchar();
    while(ch == 9 || ch == 10 || ch == 32) ch = IO :: getchar();
    return ch;
}

int n, m;
const int block = 5000;
int a[300005], bel[300005];

struct QUEST{
    int l, r;
    int t, cc;
    int id;
    inline bool operator <(const QUEST &tmp) const{
        return (bel[l] ^ bel[tmp.l]) ?(l < tmp.l) : ((bel[r] ^ bel[tmp.r])? ((bel[l] & 1)? (r < tmp.r) : (r > tmp.r)) : (t < tmp.t));
    }
}q[300005];

int c[300005];

int ans, cnt[300005];
int ret[300005];
int main(){
    read(n);read(m);
    for (register int i = 1;i <= n;i ++) read(a[i]), bel[i] = ((i - 1) / block) + 1;
    int tot = 0, tt = 0;
    for (register int i = 1;i <= m;++ i){
        int opt;read(opt);
        if(opt == 1){
            read(q[++ tot].l);read(q[tot].r);read(q[tot].cc);
            q[tot].t = tt;
            q[tot].id = tot;
        }
        else{read(c[++ tt]);}
    }
    sort(q + 1, q + tot + 1);
    int l = q[1].l, r = l - 1, t = 0;
    for (register int i = 1;i <= tot;++ i){
        while(l > q[i].l) ++ cnt[a[-- l]];
        while(r < q[i].r) ++ cnt[a[++ r]];
        while(l < q[i].l) -- cnt[a[l ++]];
        while(r > q[i].r) -- cnt[a[r --]];
        while(t < q[i].t){
            t ++;
            bool ck1 = (c[t] <= q[i].r && c[t] >= q[i].l), ck2 = (c[t] + 1 <= q[i].r && c[t] + 1 >= q[i].l);
            if(ck1 ^ ck2) {
                if(ck1) -- cnt[a[c[t]]], ++ cnt[a[c[t] + 1]];
                else ++ cnt[a[c[t]]], -- cnt[a[c[t] + 1]];
            }
            swap(a[c[t]], a[c[t] + 1]);
        }
        while(t > q[i].t){
            bool ck1 = (c[t] <= q[i].r && c[t] >= q[i].l), ck2 = (c[t] + 1 <= q[i].r && c[t] + 1 >= q[i].l);
            if(ck1 ^ ck2) {
                if(ck1) -- cnt[a[c[t]]], ++ cnt[a[c[t] + 1]];
                else ++ cnt[a[c[t]]], -- cnt[a[c[t] + 1]];
            }
            swap(a[c[t]], a[c[t] + 1]);
            t --;
        }
        ret[q[i].id] = cnt[q[i].cc];
    }
    for (register int i = 1;i <= tot;++ i) fprint(ret[i], 10), IO :: check();
    IO :: flush();
    return 0;
}