Problem:
Score : 1700 points

Problem Statement

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.

Notes

The inversion number of a sequence Z whose length N is the number of pairs of integers i and j ( 1 \leq i < j \leq N ) such that Z_i > Z_j.

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.

Sample Input 1

3
2 3 3

Sample Output 1

4

There are four permutations that satisfy the condition: (1,2,3), (1,3,2), (2,1,3) and (2,3,1).
The inversion numbers of these permutations are 0, 1, 1 and 2, respectively, for a total of 4.

Sample Input 2

6
4 2 5 1 6 3

Sample Output 2

7

Only one permutation (4,2,5,1,6,3) satisfies the condition.
The inversion number of this permutation is 7, so the answer is 7.

Sample Input 3

5
4 4 4 4 4

Sample Output 3

0

No permutation satisfies the condition.

Sample Input 4

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

Sample Output 4

848414012

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

typedef long long int int64;
static const int64 mod = 1000000007LL;

int64 exgcd(int64 a,int64 b,int64 &c,int64 &d)
{
	if(b == 0) {
		c = 1, d = 0;
		return a;
	}
	int64 x = exgcd(b,a % b,d,c);
	d -= c * (a / b);
	return a;
}
int64 finv(int64 x)
{
	int64 c,d;
	exgcd(x,mod,c,d);
	if(c < 0) c += mod;
	return c;
}

int a[200003];
int n;
int cnt[200003];
struct pair {
	int a,b;
	pair():a(0),b(0){}
	pair(int a,int b):a(a),b(b){}
	pair operator +(const pair &c) const {
		if(b > c.b || c.a == 0) return *this;
		if(c.b > b || a == 0) return c;
		return pair((a+c.a)%mod,b);
	}
	pair operator *(const pair &c) const {
		return pair((int64)a*c.a%mod,b+c.b);
	}
} arr[200003];
void add(int i,pair v)
{
	while(i <= n) {
		arr[i] = arr[i] + v;
		i += i&-i;
	}
}
pair sum(int i)
{
	pair ans = arr[i];i -= i&-i;
	while(i > 0) {
		ans = ans + arr[i];
		i -= i&-i;
	}
	return ans;
}
int arr2[200003];
void add2(int i,int v)
{
	while(i <= n) {
		arr2[i] = arr2[i] + v;
		i += i&-i;
	}
}
int sum2(int i)
{
	int ans = 0;
	while(i > 0) {
		ans += arr2[i];
		i -= i&-i;
	}
	return ans;
}
int64 inv[200003];
std::pair<pair,pair> calpair(int k)
{
	pair a = k == 0 ? pair(1,-1) : pair(k,0);
	pair b = k-1 == 0 ? pair(1,-1) : pair(k-1,0);
	return std::make_pair(pair(a.a*inv[b.a]%mod,a.b-b.b),pair(b.a*inv[a.a]%mod,b.b-a.b));
}
pair suf1[200003],suf2[200003]; // i/(i-1),(i-1)/i
pair pre1[200003],pre2[200003]; // i/(i-1),(i-1)/i
int main()
{
	scanf("%d",&n);
	inv[1] = 1;for(int i = 2;i <= n;++i) inv[i] = (mod-mod/i)*inv[mod%i]%mod;
	for(int i = 0;i < n;++i) {
		scanf("%d",&a[i]);--a[i];
		++cnt[a[i]];
	}
	for(int i = n-2;i >= 0;--i) {
		cnt[i] += cnt[i+1]-1;
	}
	int64 ians = 1;
	for(int i = 0;i < n;++i) {
		if(cnt[i] <= 0) {
			printf("0\n");
			return 0;
		}
		ians = ians * cnt[i] % mod;
	}
	for(int i = n-1;i >= 0;--i) {
		std::pair<pair,pair> p = calpair(cnt[i]);
		if(i == n-1) {
			suf1[i] = p.first;
			suf2[i] = p.second;
		} else {
			suf1[i] = p.first * suf1[i+1];
			suf2[i] = p.second * suf2[i+1];
		}
	}
	for(int i = 0;i < n;++i) {
		std::pair<pair,pair> p = calpair(cnt[i]);
		if(i == 0) {
			pre1[i] = p.first;
			pre2[i] = p.second;
		} else {
			pre1[i] = p.first * pre1[i-1];
			pre2[i] = p.second * pre2[i-1];
		}
	}
	/*
	for(int i = 0;i < n;++i) {
		printf("%d ", cnt[i]);
	}
	printf("\n");
	for(int i = 0;i < n;++i) {
		printf("%d;%d ", pre1[i].a, pre1[i].b);
	}
	printf("\n");
	for(int i = 0;i < n;++i) {
		printf("%d;%d ", pre2[i].a, pre2[i].b);
	}
	printf("\n");
	*/
	int64 ans = 0;
	// a[i] <= a[j] case
	for(int i = 0;i < n;++i) {
		pair val = sum(a[i]+1) * pre2[a[i]];
		assert(val.b <= 0);
		if(val.b == 0) ans += val.a;
		add(a[i]+1,pre1[a[i]]);
	}
	// a[i] > a[j] case
	memset(arr,0,sizeof(arr));
	int64 count = 0;
	for(int i = n-1;i >= 0;--i) {
		pair val = sum(a[i]) * pre2[a[i]];
		count += sum2(a[i]);
		assert(val.a == 0 || val.b <= 0);
		if(val.b == 0) ans -= val.a;
		add(a[i]+1,pre1[a[i]]);
		add2(a[i]+1,1);
	}
	ans = ans % mod * inv[2] % mod;
	ans = ians * (ans + count % mod) % mod;
	if(ans < 0) ans += mod;
	printf("%lld\n", ans);
}