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
#include<bits/stdc++.h>
#define ld double
#define ull unsigned long long
#define ll long long
#define pii pair<int,int >
#define iiii pair<int,pii >
#define mp make_pair
#define INF 1000000000
#define MOD 1000000007
#define rep(i,x) for(int (i)=0;(i)<(x);(i)++)
inline int getint(){
	int x=0,p=1;char c=getchar();
	while (c<=32)c=getchar();
	if(c==45)p=-p,c=getchar();
	while (c>32)x=x*10+c-48,c=getchar();
	return x*p;
}
using namespace std;
//ruogu_alter
const int N=2e5+5;
const int inv2=(MOD+1)/2;
int n,a[N],b[N],cnt[N],ib[N],pre[N],dx[N],dy[N];
//
inline void add(int &x,int y){
	x+=y;if(x>=MOD)x-=MOD;
}
inline void sub(int &x,int y){
	x-=y;if(x<0)x+=MOD;
}
inline int sub2(int x,int y){
	x-=y;
	return (x<0)?x+MOD:x;
}
inline int mul(int x,int y){
	ll ans=1ll*x*y;
	return ans%MOD;
}
inline int modpow(int x,int y){
	int ans=1;
	while(y){
		if(y&1)ans=mul(ans,x);
		x=mul(x,x);
		y>>=1;
	}
	return ans;
}
inline int modinv(int x){
	return modpow(x,MOD-2);
}
inline void upd(int x,int y){
	while(x<N){
		add(dy[x],y);
		dx[x]++;
		x+=x&-x;
	}
}	
inline int qry(int *d,int x){
	int ans=0;
	while(x){
		add(ans,d[x]);
		x-=x&-x;
	}
	return ans;
}
int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	n=getint();
	for(int i=1;i<=n;i++){
		a[i]=getint();
		cnt[a[i]]++;
	}
	for(int i=n-1;i>=1;i--)cnt[i]+=cnt[i+1];
	for(int i=1;i<=n;i++)cnt[i]-=n-i;
	for(int i=1;i<=n;i++)if(cnt[i]<=0){
		return !printf("%d\n",0);
	}
	int s=1;pre[0]=b[0]=1;
	for(int i=1;i<=n;i++)s=mul(s,cnt[i]);
	for(int i=1;i<=n;i++){
		b[i]=mul(cnt[i]-1,modinv(cnt[i]));
		if(!b[i])pre[i]=i,b[i]=b[i-1];
		else pre[i]=pre[i-1],b[i]=mul(b[i-1],b[i]);
		ib[i]=modinv(b[i]);
	}
	int res=0;
	for(int i=1;i<=n;i++){
		add(res,mul(b[a[i]],mul(s,mul(inv2,sub2(qry(dy,a[i]),qry(dy,pre[a[i]]-1))))));	
//		cout<<res<<endl;
		upd(a[i],ib[a[i]]);
	}
//	cout<<res<<endl;
	memset(dx,0,sizeof(dx));
	memset(dy,0,sizeof(dy));
	for(int i=n;i>=1;i--){
		sub(res,mul(b[a[i]],mul(s,mul(inv2,sub2(qry(dy,a[i]-1),qry(dy,pre[a[i]]-1))))));
		add(res,mul(s,qry(dx,a[i]-1)));
		upd(a[i],ib[a[i]]);
	}
	cout<<res<<endl;
	return 0;
}