Problem: AquaMoon has three integer arrays a, b, c of length n, where 1 ≤ a_i, b_i, c_i ≤ n for all i.

In order to accelerate her potato farming, she organizes her farm in a manner based on these three arrays. She is now going to complete m operations to count how many potatoes she can get. Each operation will have one of the two types:

  1. AquaMoon reorganizes their farm and makes the k-th element of the array a equal to x. In other words, perform the assignment a_k := x. 
  2. Given a positive integer r, AquaMoon receives a potato for each triplet (i,j,k), such that 1≤ i<j<k≤ r, and b_{a_i}=a_j=c_{a_k}. Count the number of such triplets. 



As AquaMoon is busy finding the library, help her complete all of their operations.

Input

The first line contains two integers n, m (1≤ n≤ 2⋅10^5, 1≤ m≤ 5⋅10^4).

The second line contains n integers a_1, a_2, ...,a_n (1≤ a_i≤ n).

The third line contains n integers b_1, b_2, ...,b_n (1≤ b_i≤ n).

The fourth line contains n integers c_1, c_2, ...,c_n (1≤ c_i≤ n).

The next m lines describe operations, the i-th line describes the i-th operation in one of these two formats:

  * "1\ k\ x" (1≤ k,x≤ n), representing an operation of the first type. 
  * "2\ r" (1≤ r≤ n), representing an operation of the second type. 



It is guaranteed that there is at least one operation of the second type.

Output

For each operation of the second type print the answer.

Example

Input


5 4
1 2 3 4 5
2 3 4 5 1
5 1 2 3 4
2 5
1 2 3
2 4
2 5


Output


3
0
2

Note

For the first operation, the triplets are:

  * i=1, j=2, k=3 
  * i=2, j=3, k=4 
  * i=3, j=4, k=5 



There is no satisfying triplet for the third operation.

For the fourth operation, the triplets are:

  * i=2, j=4, k=5 
  * i=3, j=4, k=5 

Solution (in CPP):

Solution:
//And in that light,I find deliverance.
//This code is written by Yiran Empty.
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#include<bits/stdc++.h>
#undef local
#define ll long long
using namespace std;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
int n=read(),m=read(),k;
int a[200003],b[200003],c[200003];
int B[200003],C[200003],cnt[200003],tnc[200003],ntc[200003];
int tv[200003];
ll v[200003],pre[200003],sum[200003];
const int S=230;
int op[233],x[233],y[233];
int f[200003];
//这个位置是否可变
int pos[233],len,qs;
//可变位置集合，询问个数
int tA[233],tB[233],tC[233],r[233];
int ta[233][233],tb[233][233],tc[233][233],tf[233][233];
ll ans[233];
//每个询问位置对应值的集合
struct node{int id,val,mul;};
vector<node> qb[200003],qc[200003],qd[200003],qp[200003],qA[200003],qC[200003];
#ifdef local
vector<node> qcc[200003];
#endif
int ca[200003],cb[200003],cc[200003],cd[200003];
int resA[100003],resC[100003],resid[100003],rescnt;
inline void make_query(int id,int val,int l,int r)
{
	(l<r)&&(
	qp[r].push_back((node){id,val,1}),qp[l-1].push_back((node){id,val,-1}),
	++rescnt,resid[rescnt]=id,resA[rescnt]=resC[rescnt]=0,
	qA[l-1].push_back((node){rescnt,val,1}),
	qC[r].push_back((node){rescnt,val,1}),qC[l-1].push_back((node){rescnt,val,-1}),0);
	return ;
}
#ifdef local
ll Ans[233][8];
#endif
int qwq[200003];
void solve()
{
#ifdef local
	memset(Ans,0,sizeof(Ans));
#endif
	memset(qwq,0,sizeof(qwq));
	//Init
	memset(f,0,sizeof(f)),len=0,rescnt=0,qs=0;
	int tmp=0;
	for(int i=1; i<=k; ++i) (op[i]==1&&!f[x[i]])&&(f[x[i]]=1,pos[++len]=x[i]);
	for(int i=1; i<=n; ++i) (f[i])&&(f[i]=++tmp);
	sort(pos+1,pos+len+1);
	for(int i=1; i<=len; ++i) tA[i]=a[pos[i]],tB[i]=B[pos[i]],tC[i]=C[pos[i]];
	for(int i=1; i<=k; ++i) 
		if(op[i]==1) tA[f[x[i]]]=y[i],tB[f[x[i]]]=b[y[i]],tC[f[x[i]]]=c[y[i]];
		else {r[++qs]=x[i],ans[qs]=0;for(int j=1; j<=len; ++j)ta[qs][j]=tA[j],tb[qs][j]=tB[j],tc[qs][j]=tC[j];}
	//Case 1 S S S
	//Case 2 S S D
	//Case 3 S D S (Queries making)
	for(int i=1; i<=n; ++i) 
#ifdef local
		qp[i].clear(),
#endif
		qb[i].clear(),qc[i].clear(),qd[i].clear(),qp[i].clear(),qA[i].clear(),qC[i].clear(); 
	memset(cnt,0,sizeof(cnt)),memset(sum,0,sizeof(sum));
	for(int i=1,id=0; i<=n; ++i) 
	{
		pre[i]=pre[i-1];
		if(f[i])
		{
			++id;
			for(int j=1; j<=qs; ++j) 
			tf[j][id]=qwq[tb[j][id]],
#ifdef local
			(r[j]>=i)&&(qcc[r[j]].push_back((node){j,ta[j][id],cnt[ta[j][id]]}),
			qcc[i].push_back((node){j,ta[j][id],-cnt[ta[j][id]]}),
			Ans[j][1]+=sum[tc[j][id]]
#else
			(r[j]>=i)&&(qc[r[j]].push_back((node){j,ta[j][id],cnt[ta[j][id]]}),
			qc[i].push_back((node){j,ta[j][id],-cnt[ta[j][id]]}),
			ans[j]+=sum[tc[j][id]]
#endif
			);
		}
		else v[i]=cnt[a[i]],++qwq[a[i]],++cnt[B[i]],pre[i]+=sum[C[i]],sum[a[i]]+=v[i];
	}
	for(int i=1; i<=qs; ++i) 
#ifdef local
		Ans[i][0]+=pre[r[i]];
#else
		ans[i]+=pre[r[i]];
#endif
	//Case 4 D D D
	//Case 5 D D S (Queries making)
	//Case 6 S D D (Queries making)
	//Case 7 D S D (Part 1)
	//Case 7 D S D (Part 2 queries making)
	//Case 8 D S S (Queries making)
	for(int i=1; i<=qs; ++i) 
	{
		for(int j=1; pos[j]<=r[i]&&j<=len; ++j) 
			cnt[ta[i][j]]=cnt[tb[i][j]]=0,
			sum[ta[i][j]]=sum[tc[i][j]]=0,
			ntc[tb[i][j]]=ntc[tc[i][j]]=0,
			tnc[tb[i][j]]=tnc[tc[i][j]]=0;
		int R=0;
		for(int j=1; pos[j]<=r[i]&&j<=len; ++j) 
			make_query(i,tb[i][j],pos[j]+1,r[i]),
			R=j,tv[j]=cnt[ta[i][j]],
			(ntc[tc[i][j]])&&(qd[pos[j]].push_back((node){i,tc[i][j],ntc[tc[i][j]]}),0),
			(tv[j])&&(qc[r[i]].push_back((node){i,ta[i][j],tv[j]}),
			qc[pos[j]].push_back((node){i,ta[i][j],-tv[j]}),0),
#ifdef local
			Ans[i][6]-=tnc[tc[i][j]],
			Ans[i][3]+=sum[tc[i][j]],
#else
			ans[i]-=tnc[tc[i][j]],
			ans[i]+=sum[tc[i][j]],
#endif
			tnc[tb[i][j]]+=tf[i][j],
			++cnt[tb[i][j]],++ntc[tb[i][j]],
			sum[ta[i][j]]+=tv[j];
		for(int j=1; j<=R; ++j) cnt[ta[i][j]]=cnt[tc[i][j]]=0;
		for(int j=R; j>=1; --j) 
			(cnt[ta[i][j]])&&(qb[pos[j]-1].push_back((node){i,ta[i][j],cnt[ta[i][j]]}),0),
			++cnt[tc[i][j]];
	}
	//Case 3 S D S (Queries answering)
	//Case 5 D D S (Queries answering)
	//Case 6 S D D (Queries answering)
	//Case 7 D S D (Part 2 queries answering)
	//Case 8 D S S (Queries answering)
	memset(ca,0,sizeof(ca)),memset(cb,0,sizeof(cb)),
	memset(cc,0,sizeof(cc)),memset(cd,0,sizeof(cd));
	memset(sum,0,sizeof(sum));
	for(int i=1; i<=n; ++i)
	{
		(!f[i])&&(sum[C[i]]+=ca[C[i]],++ca[a[i]],++cb[B[i]],++cc[C[i]]);
#ifdef local
		for(auto j:qb[i]) Ans[j.id][5]+=1ll*cb[j.val]*j.mul;
		for(auto j:qc[i]) Ans[j.id][4]+=1ll*cc[j.val]*j.mul;
		for(auto j:qd[i]) Ans[j.id][6]+=1ll*ca[j.val]*j.mul;
		for(auto j:qp[i]) Ans[j.id][7]+=1ll*sum[j.val]*j.mul;
		for(auto j:qcc[i])Ans[j.id][2]+=1ll*cc[j.val]*j.mul;
#else
		for(auto j:qb[i]) ans[j.id]+=1ll*cb[j.val]*j.mul;
		for(auto j:qc[i]) ans[j.id]+=1ll*cc[j.val]*j.mul;
		for(auto j:qd[i]) ans[j.id]+=1ll*ca[j.val]*j.mul;
		for(auto j:qp[i]) ans[j.id]+=1ll*sum[j.val]*j.mul;
#endif
		for(auto j:qA[i]) resA[j.id]+=ca[j.val]*j.mul;
		for(auto j:qC[i]) resC[j.id]+=cc[j.val]*j.mul;
	}
#ifdef local
	for(int i=1; i<=rescnt; ++i) Ans[resid[i]][7]-=1ll*resA[i]*resC[i];
#else
	for(int i=1; i<=rescnt; ++i) ans[resid[i]]-=1ll*resA[i]*resC[i];
#endif
	//Update & output
	for(int i=1; i<=k; ++i) (op[i]==1)&&(a[x[i]]=y[i],B[x[i]]=b[y[i]],C[x[i]]=c[y[i]]);
#ifdef local
	for(int i=1; i<=qs; ++i,puts("")) 
	for(int j=0; j<8; ++j) printf("%lld ",Ans[i][j]);
#else
	for(int i=1; i<=qs; ++i) printf("%lld\n",ans[i]);
#endif
	return ;
}
signed main()
{
	for(int i=1; i<=n; ++i) a[i]=read();
	for(int i=1; i<=n; ++i) b[i]=read();
	for(int i=1; i<=n; ++i) c[i]=read();
	for(int i=1; i<=n; ++i) B[i]=b[a[i]],C[i]=c[a[i]];
	for(int i=1; i<=m; ++i) op[++k]=read(),x[k]=read(),(op[k]==1)&&(y[k]=read()),(k==S)&&(solve(),k=0);
	if(k) solve();
	return 0;
}