#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef vector<int> vi;
typedef pair<int,int> pii;
const int N=200000+100;
const db pi=acos(-1.0);
#define lowbit(x) ((x)&(-x))
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u],v=sq[i].to;i;i=sq[i].nxt,v=sq[i].to)
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define maxd 1000000007
#define inv2 500000004
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

inline ll readll()
{
    ll x=0;int f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

namespace My_Math{
    #define N 100000

    int fac[N+100],invfac[N+100];

    int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
    int dec(int x,int y) {return x<y?x-y+maxd:x-y;}
    int mul(int x,int y) {return 1ll*x*y%maxd;}
    ll qpow(ll x,int y)
    {
        ll ans=1;
        while (y)
        {
            if (y&1) ans=mul(ans,x);
            x=mul(x,x);y>>=1;
        }
        return ans;
    }
    int getinv(int x) {return qpow(x,maxd-2);}

    int C(int n,int m)
    {
        if ((n<m) || (n<0) || (m<0)) return 0;
        return mul(mul(fac[n],invfac[m]),invfac[n-m]);
    }

    void math_init()
    {
        fac[0]=invfac[0]=1;
        rep(i,1,N) fac[i]=mul(fac[i-1],i);
        invfac[N]=getinv(fac[N]);
        per(i,N-1,1) invfac[i]=mul(invfac[i+1],i+1);
    }
    #undef N
}
using namespace My_Math;

int n,a[N],c[N],d[N],pos0[N],invd[N],cnt0[N];

struct Fenwick_Tree{

    int tr[N];

    void clr()
    {
        rep(i,1,n) tr[i]=0;
    }

    void modify(int p,int x)
    {
        for (int i=p;i<=n;i+=lowbit(i)) tr[i]=add(tr[i],x);
    }

    int query(int p)
    {
        int ans=0;
        for (int i=p;i;i-=lowbit(i)) ans=add(ans,tr[i]);
        return ans;
    }
}tr0,tr1;

int main()
{
    n=read();
    rep(i,1,n) 
    {
        a[i]=read();c[a[i]]++;
    }
    per(i,n,1) c[i]+=c[i+1];
    int sum=1;
    rep(i,1,n)
    {
        if (c[i]<n-i) {puts("0");return 0;}
        c[i]-=(n-i);
        sum=mul(sum,c[i]);
    }
    //cout << sum << endl;
    d[0]=1;pos0[0]=1;
    rep(i,1,n)
    {
        int x=mul(c[i]-1,getinv(c[i]));cnt0[i]=cnt0[i-1];
        if (!x)
        {
            cnt0[i]++;pos0[cnt0[i]]=i;
            d[i]=d[i-1];
        }
        else
        {
            d[i]=mul(d[i-1],x);
        }
        invd[i]=getinv(d[i]);
    }
    //rep(i,1,n) cout << d[i] << " ";cout << endl;
    int ans=0;
    rep(i,1,n)
    {
        int tmp=dec(tr1.query(a[i]),tr1.query(pos0[cnt0[a[i]]]-1));
        ans=add(ans,mul(mul(d[a[i]],tmp),mul(sum,inv2)));
        tr1.modify(a[i],invd[a[i]]);
    }
    tr1.clr();
    per(i,n,1)
    {
        int tmp=dec(tr1.query(a[i]-1),tr1.query(pos0[cnt0[a[i]]]-1));
        ans=dec(ans,mul(mul(d[a[i]],tmp),mul(sum,inv2)));
        tmp=tr0.query(a[i]-1);
        ans=add(ans,mul(tmp,sum));
        tr1.modify(a[i],invd[a[i]]);
        tr0.modify(a[i],1);
    }
    printf("%d\n",ans);
    return 0;
}
        


### Prompt

Develop a solution in Cpp 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
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef vector<int> vi;
typedef pair<int,int> pii;
const int N=200000+100;
const db pi=acos(-1.0);
#define lowbit(x) ((x)&(-x))
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u],v=sq[i].to;i;i=sq[i].nxt,v=sq[i].to)
#define fir first
#define sec second
#define mkp make_pair
#define pb push_back
#define maxd 1000000007
#define inv2 500000004
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

inline ll readll()
{
    ll x=0;int f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

namespace My_Math{
    #define N 100000

    int fac[N+100],invfac[N+100];

    int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
    int dec(int x,int y) {return x<y?x-y+maxd:x-y;}
    int mul(int x,int y) {return 1ll*x*y%maxd;}
    ll qpow(ll x,int y)
    {
        ll ans=1;
        while (y)
        {
            if (y&1) ans=mul(ans,x);
            x=mul(x,x);y>>=1;
        }
        return ans;
    }
    int getinv(int x) {return qpow(x,maxd-2);}

    int C(int n,int m)
    {
        if ((n<m) || (n<0) || (m<0)) return 0;
        return mul(mul(fac[n],invfac[m]),invfac[n-m]);
    }

    void math_init()
    {
        fac[0]=invfac[0]=1;
        rep(i,1,N) fac[i]=mul(fac[i-1],i);
        invfac[N]=getinv(fac[N]);
        per(i,N-1,1) invfac[i]=mul(invfac[i+1],i+1);
    }
    #undef N
}
using namespace My_Math;

int n,a[N],c[N],d[N],pos0[N],invd[N],cnt0[N];

struct Fenwick_Tree{

    int tr[N];

    void clr()
    {
        rep(i,1,n) tr[i]=0;
    }

    void modify(int p,int x)
    {
        for (int i=p;i<=n;i+=lowbit(i)) tr[i]=add(tr[i],x);
    }

    int query(int p)
    {
        int ans=0;
        for (int i=p;i;i-=lowbit(i)) ans=add(ans,tr[i]);
        return ans;
    }
}tr0,tr1;

int main()
{
    n=read();
    rep(i,1,n) 
    {
        a[i]=read();c[a[i]]++;
    }
    per(i,n,1) c[i]+=c[i+1];
    int sum=1;
    rep(i,1,n)
    {
        if (c[i]<n-i) {puts("0");return 0;}
        c[i]-=(n-i);
        sum=mul(sum,c[i]);
    }
    //cout << sum << endl;
    d[0]=1;pos0[0]=1;
    rep(i,1,n)
    {
        int x=mul(c[i]-1,getinv(c[i]));cnt0[i]=cnt0[i-1];
        if (!x)
        {
            cnt0[i]++;pos0[cnt0[i]]=i;
            d[i]=d[i-1];
        }
        else
        {
            d[i]=mul(d[i-1],x);
        }
        invd[i]=getinv(d[i]);
    }
    //rep(i,1,n) cout << d[i] << " ";cout << endl;
    int ans=0;
    rep(i,1,n)
    {
        int tmp=dec(tr1.query(a[i]),tr1.query(pos0[cnt0[a[i]]]-1));
        ans=add(ans,mul(mul(d[a[i]],tmp),mul(sum,inv2)));
        tr1.modify(a[i],invd[a[i]]);
    }
    tr1.clr();
    per(i,n,1)
    {
        int tmp=dec(tr1.query(a[i]-1),tr1.query(pos0[cnt0[a[i]]]-1));
        ans=dec(ans,mul(mul(d[a[i]],tmp),mul(sum,inv2)));
        tmp=tr0.query(a[i]-1);
        ans=add(ans,mul(tmp,sum));
        tr1.modify(a[i],invd[a[i]]);
        tr0.modify(a[i],1);
    }
    printf("%d\n",ans);
    return 0;
}
        


```