//一个奇怪的做法...QwQ
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define qwq(x) cout<<"# "<<#x<<" = "<<x<<endl;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

namespace{
	const int mod=1e9+7;
	int power(int x,int p){
		int res=1;
		for(;p;p>>=1,x=(ll)x*x%mod)
			if(p&1)res=(ll)res*x%mod;
		return res;
	}
	int mul(int x,int y){
		return static_cast<ll>(x)*y%mod;
	}
	int fix(int x){
		return x>=mod?x-mod:x;
	}
	void add(int &x,int y){
		x=(x+y>mod?x+y-mod:x+y);
	}
}

const int N=200005;
int a[N],b[N],tmp[N],cur[N],n,ans;

namespace Part1{
	int f[N][5];
	int g(int x,int y){
		int res=b[x]-x+1-y;
		return res<=0?0:res;
	}
	int C2(int x){
		return (ll)x*(x-1)/2%mod;
	}
	void rmain(){
		f[0][0]=1;
		rep(i,1,n){
			REP(t,5)
				add(f[i][t],mul(f[i-1][t],g(i,2-abs(t-2))));
			add(f[i][1],mul(mul(f[i-1][0],g(i,1)),b[i]-b[i-1]));
			add(f[i][2],mul(mul(f[i-1][1],g(i,2)),b[i]-b[i-1]));
			add(f[i][2],mul(mul(f[i-1][0],g(i,2)),C2(b[i]-b[i-1])));
			add(f[i][3],f[i-1][2]);
			add(f[i][3],mul(f[i-1][1],b[i]-b[i-1]));
			add(f[i][3],mul(f[i-1][0],C2(b[i]-b[i-1])));
			add(f[i][4],f[i-1][3]);
		}
		add(ans,f[n][4]);
	}
}

namespace Part2{
	int S[N<<2],C[N<<2],s[N],pre[N],suf[N],Suf[N],las[N];
	#define lc (o<<1)
	#define rc (o<<1|1)
	#define mid ((l+r)>>1)
	void app(int o,int l,int r,int x){
		add(S[o],mul(x,s[r]-s[l-1]+mod));
		add(C[o],x);
	}
	void down(int o,int l,int r){
		if(C[o]){
			app(lc,l,mid,C[o]);
			app(rc,mid+1,r,C[o]);
			C[o]=0;
		}
	}
	void upd(int o,int l,int r,int x,int y,int z){
		if(l==x&&y==r){
			app(o,l,r,z);
			return;
		}
		down(o,l,r);
		if(x<=mid)upd(lc,l,mid,x,min(y,mid),z);
		if(mid<y)upd(rc,mid+1,r,max(mid+1,x),y,z);
		S[o]=fix(S[lc]+S[rc]);
	}
	int qry(int o,int l,int r,int x,int y){
		if(l==x&&y==r)return S[o];
		down(o,l,r);
		if(y<=mid)return qry(lc,l,mid,x,y);
		if(mid<x)return qry(rc,mid+1,r,x,y);
		return fix(qry(lc,l,mid,x,mid)+qry(rc,mid+1,r,mid+1,y));
	}
	void rmain(){
		pre[0]=1;
		rep(i,1,n)
			pre[i]=mul(pre[i-1],b[i]-i+1);
		Suf[n+1]=suf[n+1]=1;
		per(i,n,1){
			suf[i]=mul(suf[i+1],max(1,b[i]-i));
			Suf[i]=mul(Suf[i+1],b[i]-i+1);
		}
		rep(i,1,n){
			int p=lower_bound(b+1,b+n+1,i)-b;
			s[i]=mul(pre[p-1],suf[p]);
			add(s[i],s[i-1]);
//			printf("%d:  %d %d %d\n",i,s[i],pre[i],suf[i]);
		}
		rep(i,1,n)
			if(b[i]==i)las[i]=i;
			else las[i]=las[i-1];
		rep(i,1,n){
			if(a[i]<n){
				add(ans,qry(1,1,n,a[i]+1,n));
//				printf("# %d %d\n",i,qry(1,1,n,a[i]+1,n));
			}
			int coef=mul(power(suf[cur[i]],mod-2),Suf[cur[i]+1]);
			if(las[cur[i]]==cur[i]&&b[cur[i]-1]!=a[i]){
				upd(1,1,n,a[i],a[i],coef);
			}
			else{
//				printf("@ %d %d %d\n",i,b[las[cur[i]-1]],coef);
				upd(1,1,n,b[las[cur[i]-1]]+1,a[i],coef);
			}
		}
	}
}
/*
5
4 5 4 1 2 

4 5 3 1 2

3 5 4 1 2
*/
int main(){
	read(n);
	rep(i,1,n){
		read(a[i]);
		tmp[i]=i;
	}
	sort(tmp+1,tmp+n+1,[&](int x,int y){
		return a[x]<a[y];
	});
	rep(i,1,n)cur[tmp[i]]=i;
	rep(i,1,n)b[cur[i]]=a[i];
	rep(i,1,n)
		if(b[i]<i)return puts("0"),0;
	Part1::rmain();
	Part2::rmain();
	printf("%d\n",ans);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

Snuke has an integer sequence A whose length is N. He likes permutations of (1, 2, ..., N), P, that satisfy the following condition:

* P_i \leq A_i for all i ( 1 \leq i \leq N ).



Snuke is interested in the inversion numbers of such permutations. Find the sum of the inversion numbers over all permutations that satisfy the condition. Since this can be extremely large, compute the sum modulo 10^9+7.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq A_i \leq N ( 1 \leq i \leq N )
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the sum of the inversion numbers over all permutations that satisfy the condition.

Examples

Input

3
2 3 3


Output

4


Input

6
4 2 5 1 6 3


Output

7


Input

5
4 4 4 4 4


Output

0


Input

30
22 30 15 20 10 29 11 29 28 11 26 10 18 28 22 5 29 16 24 24 27 10 21 30 29 19 28 27 18 23


Output

848414012

### Response

```cpp
//一个奇怪的做法...QwQ
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define qwq(x) cout<<"# "<<#x<<" = "<<x<<endl;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

namespace{
	const int mod=1e9+7;
	int power(int x,int p){
		int res=1;
		for(;p;p>>=1,x=(ll)x*x%mod)
			if(p&1)res=(ll)res*x%mod;
		return res;
	}
	int mul(int x,int y){
		return static_cast<ll>(x)*y%mod;
	}
	int fix(int x){
		return x>=mod?x-mod:x;
	}
	void add(int &x,int y){
		x=(x+y>mod?x+y-mod:x+y);
	}
}

const int N=200005;
int a[N],b[N],tmp[N],cur[N],n,ans;

namespace Part1{
	int f[N][5];
	int g(int x,int y){
		int res=b[x]-x+1-y;
		return res<=0?0:res;
	}
	int C2(int x){
		return (ll)x*(x-1)/2%mod;
	}
	void rmain(){
		f[0][0]=1;
		rep(i,1,n){
			REP(t,5)
				add(f[i][t],mul(f[i-1][t],g(i,2-abs(t-2))));
			add(f[i][1],mul(mul(f[i-1][0],g(i,1)),b[i]-b[i-1]));
			add(f[i][2],mul(mul(f[i-1][1],g(i,2)),b[i]-b[i-1]));
			add(f[i][2],mul(mul(f[i-1][0],g(i,2)),C2(b[i]-b[i-1])));
			add(f[i][3],f[i-1][2]);
			add(f[i][3],mul(f[i-1][1],b[i]-b[i-1]));
			add(f[i][3],mul(f[i-1][0],C2(b[i]-b[i-1])));
			add(f[i][4],f[i-1][3]);
		}
		add(ans,f[n][4]);
	}
}

namespace Part2{
	int S[N<<2],C[N<<2],s[N],pre[N],suf[N],Suf[N],las[N];
	#define lc (o<<1)
	#define rc (o<<1|1)
	#define mid ((l+r)>>1)
	void app(int o,int l,int r,int x){
		add(S[o],mul(x,s[r]-s[l-1]+mod));
		add(C[o],x);
	}
	void down(int o,int l,int r){
		if(C[o]){
			app(lc,l,mid,C[o]);
			app(rc,mid+1,r,C[o]);
			C[o]=0;
		}
	}
	void upd(int o,int l,int r,int x,int y,int z){
		if(l==x&&y==r){
			app(o,l,r,z);
			return;
		}
		down(o,l,r);
		if(x<=mid)upd(lc,l,mid,x,min(y,mid),z);
		if(mid<y)upd(rc,mid+1,r,max(mid+1,x),y,z);
		S[o]=fix(S[lc]+S[rc]);
	}
	int qry(int o,int l,int r,int x,int y){
		if(l==x&&y==r)return S[o];
		down(o,l,r);
		if(y<=mid)return qry(lc,l,mid,x,y);
		if(mid<x)return qry(rc,mid+1,r,x,y);
		return fix(qry(lc,l,mid,x,mid)+qry(rc,mid+1,r,mid+1,y));
	}
	void rmain(){
		pre[0]=1;
		rep(i,1,n)
			pre[i]=mul(pre[i-1],b[i]-i+1);
		Suf[n+1]=suf[n+1]=1;
		per(i,n,1){
			suf[i]=mul(suf[i+1],max(1,b[i]-i));
			Suf[i]=mul(Suf[i+1],b[i]-i+1);
		}
		rep(i,1,n){
			int p=lower_bound(b+1,b+n+1,i)-b;
			s[i]=mul(pre[p-1],suf[p]);
			add(s[i],s[i-1]);
//			printf("%d:  %d %d %d\n",i,s[i],pre[i],suf[i]);
		}
		rep(i,1,n)
			if(b[i]==i)las[i]=i;
			else las[i]=las[i-1];
		rep(i,1,n){
			if(a[i]<n){
				add(ans,qry(1,1,n,a[i]+1,n));
//				printf("# %d %d\n",i,qry(1,1,n,a[i]+1,n));
			}
			int coef=mul(power(suf[cur[i]],mod-2),Suf[cur[i]+1]);
			if(las[cur[i]]==cur[i]&&b[cur[i]-1]!=a[i]){
				upd(1,1,n,a[i],a[i],coef);
			}
			else{
//				printf("@ %d %d %d\n",i,b[las[cur[i]-1]],coef);
				upd(1,1,n,b[las[cur[i]-1]]+1,a[i],coef);
			}
		}
	}
}
/*
5
4 5 4 1 2 

4 5 3 1 2

3 5 4 1 2
*/
int main(){
	read(n);
	rep(i,1,n){
		read(a[i]);
		tmp[i]=i;
	}
	sort(tmp+1,tmp+n+1,[&](int x,int y){
		return a[x]<a[y];
	});
	rep(i,1,n)cur[tmp[i]]=i;
	rep(i,1,n)b[cur[i]]=a[i];
	rep(i,1,n)
		if(b[i]<i)return puts("0"),0;
	Part1::rmain();
	Part2::rmain();
	printf("%d\n",ans);
	return 0;
}
```