Problem: You are given an integer sequence a of length N. How many permutations p of the integers 1 through N satisfy the following condition?

* For each 1 ≤ i ≤ N, at least one of the following holds: p_i = a_i and p_{p_i} = a_i.



Find the count modulo 10^9 + 7.

Constraints

* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ N

Input

The input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the number of the permutations p that satisfy the condition, modulo 10^9 + 7.

Examples

Input

3
1 2 3


Output

4


Input

2
1 1


Output

1


Input

3
2 1 1


Output

2


Input

3
1 1 1


Output

0


Input

13
2 1 4 3 6 7 5 9 10 8 8 9 11


Output

6

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define re(i,a,b) for(int i=(a);i<(b);i++)
#define repd(i,a,b) for(int i=(a);i>=(b);i--)
#define clr(a) memset(a,0,sizeof(a));
#define il inline
#define sz(a) ((int)o.size())
#define run(x) for(int k=head[x];k;k=e[k].ne)
#define v e[k].t
#define all(a) o.begin(),o.end()
#define mp make_pair
#define pb push_back 
#define w1 first
#define w2 second
#define adm(a,b,c) {a=a+b;if(a>=c)a-=c;else if(a<0)a+=c;}
typedef double db;
typedef long long ll;typedef long double ld;typedef unsigned long long ull;
typedef pair<int,int> pa;
const int N=1e6+5,M=1e7+5,INF=1e9,mod=1e9+7;
const ll linf=1e18;const double eps=1e-8,pi=acos(-1);
il int gmin(int &a,int b){if(a>b)a=b;}il ll gmin(ll &a,ll b){if(a>b)a=b;}il int gmax(int &a,int b){if(a<b)a=b;}il ll gmax(ll &a,ll b){if(a<b)a=b;}
il void read(ll&x){ll f=1,t=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){t=t*10+ch-'0';ch=getchar();}x=t*f;}il ll read(ll&x,ll&y){read(x);read(y);}
il void read(int&x){int f=1,t=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){t=t*10+ch-'0';ch=getchar();}x=t*f;}il int read(int&x,int&y){read(x);read(y);}
il void read(int&a,int&b,int&c){read(a);read(b);read(c);}il void read(ll&a,ll&b,ll&c){read(a);read(b);read(c);}
il int read(){int x;read(x);return x;}
il ll qpow(ll a,ll b,ll p){ll ret=1;for(;b;b>>=1,a=a*a%p)if(b&1)ret=ret*a%p;return ret;}il ll qpow(ll a,ll b){ll ret=1;for(;b;b>>=1,a=a*a%mod)if(b&1)ret=ret*a%mod;return ret;}
il ll qmul(ll a,ll b,ll p){ll ret=0;for(;b;b>>=1,a=(a<<1)%p)if(b&1)adm(ret,a,p);return ret;}il ll qmul(ll a,ll b){ll ret=0;for(;b;b>>=1,a=(a<<1)%mod)if(b&1)adm(ret,a,mod);return ret;}
il void judge(){
	freopen("dato.in","r",stdin);
	freopen("dato.out","w",stdout);
}
int n,res=1,a[N],vis[N],d[N],len[N],cnt[N];
ll f[N];
void gao(int n,int m){
	ll f1=1;if(n>1&&n%2==1)f1++;
	f[0]=1;f[1]=f1;
	rep(i,2,m)(f[i]=f1*f[i-1]%mod+f[i-2]*(i-1)%mod*n%mod)%=mod;
	res=res*f[m]%mod;
}
int main(){
	read(n);rep(i,1,n)read(a[i]),d[a[i]]++;
	rep(i,1,n)if(d[i]>2)return puts("0"),0;
	rep(i,1,n)if(!d[i]){
		int x=i,sz=0;
		for(;d[x]<2;x=a[x])vis[x]=1,sz++;
		len[x]=sz;
	}
	rep(i,1,n)if(!vis[i]){
		int x;vector<int>o;
		for(x=i;!vis[x];x=a[x])vis[x]=1,o.pb(x);
		if(x!=i)return puts("0"),0;
		int isc=1;re(j,0,o.size())if(len[o[j]])isc=0;
		if(isc)cnt[o.size()]++;else{
			int tmp=1;
			repd(j,o.size()-1,0)if(!len[o[j]])tmp++;else break;
			re(j,0,o.size())if(len[o[j]]){
				if(len[o[j]]<tmp)res=res*2%mod;
				if(len[o[j]]>tmp)return puts("0"),0;
				tmp=1;
			}else tmp++;
		}
	}rep(i,1,n)if(cnt[i])gao(i,cnt[i]);
	cout<<res<<endl;
}