#include <bits/stdc++.h>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

typedef long long cat;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

cat pw(cat a, cat e, cat mod) {
	if(e <= 0) return 1;
	cat x = pw(a, e/2, mod);
	x = x * x % mod;
	if(e%2) x = x * a % mod;
	return x;
}

struct fin {
	vector<cat> T;
	cat mod;
	fin(cat mod) : mod(mod) {}
	fin(int N, cat mod) : T(N+10, 0), mod(mod) {}

	int lastone(int x) {return x&(x^(x-1));}

	void put(int pos, cat val) {
		for(int i = pos+1; i < (int)T.size(); i += lastone(i)) {
			T[i] += val;
			if(T[i] >= mod) T[i] -= mod;
		}
	}

	cat get(int pos) {
		cat ret = 0;
		for(int i = pos+1; i > 0; i -= lastone(i)) {
			ret += T[i];
			if(ret >= mod) ret -= mod;
		}
		return ret;
	}
};

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N;
	cin >> N;
	vector<int> A(N);
	map<int, int> G;
	G[-1] = 0;
	for(int i = 0; i < N; i++) {
		cin >> A[i];
		G[--A[i]] = 0;
	}

	int g = 0;
	vector<int> group_val;
	ALL_THE(G, it) {
		it->ss = g++;
		group_val.push_back(it->ff);
	}
	if(group_val.back() != N-1) {
		cout << "0\n";
		 return 0;
	}
	g--;
	vector<int> group_size(g), group_id(N), num_pos(g, 0);
	for(int i = 0; i < N; i++) {
		group_id[i] = G[A[i]]-1;
		num_pos[group_id[i]]++;
	}
	for(int i = 0; i < g; i++) group_size[i] = group_val[i+1]-group_val[i];

	for(int i = g-2; i >= 0; i--) num_pos[i] += num_pos[i+1]-group_size[i+1];
	for(int i = 0; i < g; i++) if(num_pos[i] < group_size[i]) {
		cout << "0\n";
		return 0;
	}
	cat cnt_tot = 1, mod = 1000000007;
	for(int i = 0; i < g; i++)
		for(int j = num_pos[i]; j > num_pos[i]-group_size[i]; j--) cnt_tot = cnt_tot * j % mod;

	vector<cat> inv(N+2, 1);
	for(int i = 1; i <= N+1; i++) inv[i] = pw(i, mod-2, mod);
	
	cat ans = 0;
	vector<cat> F(g+1, 1), fzero(g+1, 0), Fi(g+1);
	for(int i = 0; i < g; i++) {
		if(num_pos[i] == group_size[i]) F[i+1] = F[i], fzero[i+1] = fzero[i] + 1;
		else F[i+1] = F[i] * (num_pos[i]-group_size[i]) % mod, fzero[i+1] = fzero[i];
		F[i+1] = F[i+1] * inv[num_pos[i]] % mod;
	}
	vector<int> minfzero(g, 0);
	for(int i = 1; i < g; i++) {
		if(fzero[i+1] == fzero[i]) minfzero[i] = minfzero[i-1];
		else minfzero[i] = i;
	}
	for(int i = 0; i <= g; i++) Fi[i] = pw(F[i], mod-2, mod);
	// left group > right group
	fin Fpairs(N, mod), Fsums(N, mod);
	for(int i = N-1; i >= 0; i--) {
		ans += Fpairs.get(group_id[i]-1);
		Fpairs.put(group_id[i], 1);
	}
	for(int i = N-1; i >= 0; i--) {
		int mfz = minfzero[group_id[i]];
		ans -= (Fsums.get(group_id[i]-1) - Fsums.get(mfz-1)) * inv[2] % mod * F[group_id[i]+1] % mod;
		Fsums.put(group_id[i], Fi[group_id[i]+1]);
	}
	// left group <= right group
	Fsums = fin(N, mod);
	for(int i = 0; i < N; i++) {
		int mfz = minfzero[group_id[i]];
		ans += (Fsums.get(group_id[i]) - Fsums.get(mfz-1)) * inv[2] % mod * F[group_id[i]+1] % mod;
		Fsums.put(group_id[i], Fi[group_id[i]+1]);
	}
	ans = ans % mod * cnt_tot % mod;
	if(ans < 0) ans += mod;
	cout << ans << "\n";
	return 0;
}

// look at my code
// my code is amazing

### Prompt

Your task is to create a cpp solution to the following problem:

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 <bits/stdc++.h>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

typedef long long cat;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

cat pw(cat a, cat e, cat mod) {
	if(e <= 0) return 1;
	cat x = pw(a, e/2, mod);
	x = x * x % mod;
	if(e%2) x = x * a % mod;
	return x;
}

struct fin {
	vector<cat> T;
	cat mod;
	fin(cat mod) : mod(mod) {}
	fin(int N, cat mod) : T(N+10, 0), mod(mod) {}

	int lastone(int x) {return x&(x^(x-1));}

	void put(int pos, cat val) {
		for(int i = pos+1; i < (int)T.size(); i += lastone(i)) {
			T[i] += val;
			if(T[i] >= mod) T[i] -= mod;
		}
	}

	cat get(int pos) {
		cat ret = 0;
		for(int i = pos+1; i > 0; i -= lastone(i)) {
			ret += T[i];
			if(ret >= mod) ret -= mod;
		}
		return ret;
	}
};

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N;
	cin >> N;
	vector<int> A(N);
	map<int, int> G;
	G[-1] = 0;
	for(int i = 0; i < N; i++) {
		cin >> A[i];
		G[--A[i]] = 0;
	}

	int g = 0;
	vector<int> group_val;
	ALL_THE(G, it) {
		it->ss = g++;
		group_val.push_back(it->ff);
	}
	if(group_val.back() != N-1) {
		cout << "0\n";
		 return 0;
	}
	g--;
	vector<int> group_size(g), group_id(N), num_pos(g, 0);
	for(int i = 0; i < N; i++) {
		group_id[i] = G[A[i]]-1;
		num_pos[group_id[i]]++;
	}
	for(int i = 0; i < g; i++) group_size[i] = group_val[i+1]-group_val[i];

	for(int i = g-2; i >= 0; i--) num_pos[i] += num_pos[i+1]-group_size[i+1];
	for(int i = 0; i < g; i++) if(num_pos[i] < group_size[i]) {
		cout << "0\n";
		return 0;
	}
	cat cnt_tot = 1, mod = 1000000007;
	for(int i = 0; i < g; i++)
		for(int j = num_pos[i]; j > num_pos[i]-group_size[i]; j--) cnt_tot = cnt_tot * j % mod;

	vector<cat> inv(N+2, 1);
	for(int i = 1; i <= N+1; i++) inv[i] = pw(i, mod-2, mod);
	
	cat ans = 0;
	vector<cat> F(g+1, 1), fzero(g+1, 0), Fi(g+1);
	for(int i = 0; i < g; i++) {
		if(num_pos[i] == group_size[i]) F[i+1] = F[i], fzero[i+1] = fzero[i] + 1;
		else F[i+1] = F[i] * (num_pos[i]-group_size[i]) % mod, fzero[i+1] = fzero[i];
		F[i+1] = F[i+1] * inv[num_pos[i]] % mod;
	}
	vector<int> minfzero(g, 0);
	for(int i = 1; i < g; i++) {
		if(fzero[i+1] == fzero[i]) minfzero[i] = minfzero[i-1];
		else minfzero[i] = i;
	}
	for(int i = 0; i <= g; i++) Fi[i] = pw(F[i], mod-2, mod);
	// left group > right group
	fin Fpairs(N, mod), Fsums(N, mod);
	for(int i = N-1; i >= 0; i--) {
		ans += Fpairs.get(group_id[i]-1);
		Fpairs.put(group_id[i], 1);
	}
	for(int i = N-1; i >= 0; i--) {
		int mfz = minfzero[group_id[i]];
		ans -= (Fsums.get(group_id[i]-1) - Fsums.get(mfz-1)) * inv[2] % mod * F[group_id[i]+1] % mod;
		Fsums.put(group_id[i], Fi[group_id[i]+1]);
	}
	// left group <= right group
	Fsums = fin(N, mod);
	for(int i = 0; i < N; i++) {
		int mfz = minfzero[group_id[i]];
		ans += (Fsums.get(group_id[i]) - Fsums.get(mfz-1)) * inv[2] % mod * F[group_id[i]+1] % mod;
		Fsums.put(group_id[i], Fi[group_id[i]+1]);
	}
	ans = ans % mod * cnt_tot % mod;
	if(ans < 0) ans += mod;
	cout << ans << "\n";
	return 0;
}

// look at my code
// my code is amazing

```