
#include <cstdio>
#include <cstring>

using namespace std;
const int Max_N(200050);
const int MOD(1000000000 + 7);

namespace io
{
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	inline void flush (){fwrite (obuf, 1, oS - obuf, stdout);oS = obuf;}
	inline void putc (char x){*oS ++ = x;if (oS == oT) flush ();}
	template <class I>
	inline void gi (I &x) {for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;}
	template <class I>
	inline void print (I x){
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;while(x) qu[++ qr] = x % 10 + '0',  x /= 10;while (qr) putc (qu[qr--]);}
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io::gi;
using io::putc;
using io::print;

constexpr int Mult(int a, int b)
{
	return a * 1LL * b % MOD;
}

constexpr int Add(int a, int b)
{
	return a + b >= MOD ? a + b - MOD : a + b;
}

constexpr int Sub(int a, int b)
{
	return a - b < 0 ? a - b + MOD : a - b;
}

void exgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
		x = 1, y = 0;
	else
		exgcd(b, a % b, y, x), y -= x * (a / b);
}

inline int inverse(int a)
{
	int invx, invy;
	exgcd(a, MOD, invx, invy);
	return (invx % MOD + MOD) % MOD;
}

int N, A[Max_N], C[Max_N], Fac[Max_N], Inv[Max_N], Ans, Cnt, inv2, Old[Max_N], Newa[Max_N], Newb[Max_N], Newm[Max_N], Pre[Max_N], Lst0[Max_N];

inline int work(int l, int r)
{
	return l > r ? 1 : (1 <= l && r <= N ? Mult(Fac[r], Inv[l - 1]) : 0);
}

int BIT[Max_N];

constexpr int lowbit(int x)
{
	return x & -x;
}

void insert(int i, int v)
{
	while (i)
		BIT[i] = Add(BIT[i], v), i -= lowbit(i);
}

int query(int i)
{
	int Ret(0);
	while (i <= N)
		Ret = Add(Ret, BIT[i]), i += lowbit(i);
	return Ret;
}

int main()
{
	inv2 = inverse(2);
	gi(N);
	for (int i = 1;i <= N;++i)
		gi(A[i]), ++C[A[i]];
	for (int x = 1;x <= N;++x)
		C[x] += C[x - 1];
	Fac[0] = 1;
	for (int i = 1;i <= N;++i)
		Fac[i] = Mult(Fac[i - 1], i);
	Inv[N] = inverse(Fac[N]);
	for (int i = N - 1;i >= 0;--i)
		Inv[i] = Mult(Inv[i + 1], i + 1);
	Cnt = 1;
	Pre[0] = 1;
	for (int x = 1;x <= N;++x)
	{
		Old[x] = work(x - C[x] + 1, x - C[x - 1]);
		if (Old[x] == 0)
		{
			printf("0");
			return 0;
		}
		Cnt = Mult(Cnt, Old[x]);
		Newa[x] = Mult(inverse(Old[x]), work(x - C[x], x - C[x - 1]));
		Newb[x] = Mult(inverse(Old[x]), work(x - C[x] + 1, x - C[x - 1] - 1));
		Newm[x] = Mult(inverse(Old[x]), work(x - C[x], x - C[x - 1] - 1));
		Lst0[x] = (Newm[x] == 0 ? x : Lst0[x - 1]);
		Pre[x] = Mult(Pre[x - 1], Newm[x] == 0 ? 1 : Newm[x]);
	}
	for (int x = 1;x <= N;++x)
		Ans = Add(Ans, Mult(Mult(Mult((C[x] - C[x - 1]), (C[x] - C[x - 1]) - 1), inv2), Mult(Cnt, inv2)));
	for (int i = 1, j;i <= N;++i)
	{
		j = Lst0[A[i] - 1];
		if (j == 0)
			j = 1;
		Ans = Add(Ans, Mult(inv2, Mult(Cnt, Mult(Sub(query(j), query(A[i])), Mult(Newb[A[i]], Pre[A[i] - 1])))));
		insert(A[i], Mult(Newa[A[i]], inverse(Pre[A[i]])));
	}
	memset(BIT, 0, sizeof(BIT));
	for (int i = N, j;i >= 1;--i)
	{
		j = Lst0[A[i] - 1];
		if (j == 0)
			j = 1;
		Ans = Sub(Ans, Mult(inv2, Mult(Cnt, Mult(Sub(query(j), query(A[i])), Mult(Newb[A[i]], Pre[A[i] - 1])))));
		insert(A[i], Mult(Newa[A[i]], inverse(Pre[A[i]])));
	}
	memset(BIT, 0, sizeof(BIT));
	for (int i = N, j;i >= 1;--i)
	{
		Ans = Add(Ans, Mult(Cnt, Sub(query(1), query(A[i]))));
		insert(A[i], 1);
	}
	print(Ans);
	return 0;
}
### Prompt

Please create a solution in Cpp 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 <cstdio>
#include <cstring>

using namespace std;
const int Max_N(200050);
const int MOD(1000000000 + 7);

namespace io
{
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	inline void flush (){fwrite (obuf, 1, oS - obuf, stdout);oS = obuf;}
	inline void putc (char x){*oS ++ = x;if (oS == oT) flush ();}
	template <class I>
	inline void gi (I &x) {for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;}
	template <class I>
	inline void print (I x){
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;while(x) qu[++ qr] = x % 10 + '0',  x /= 10;while (qr) putc (qu[qr--]);}
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io::gi;
using io::putc;
using io::print;

constexpr int Mult(int a, int b)
{
	return a * 1LL * b % MOD;
}

constexpr int Add(int a, int b)
{
	return a + b >= MOD ? a + b - MOD : a + b;
}

constexpr int Sub(int a, int b)
{
	return a - b < 0 ? a - b + MOD : a - b;
}

void exgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
		x = 1, y = 0;
	else
		exgcd(b, a % b, y, x), y -= x * (a / b);
}

inline int inverse(int a)
{
	int invx, invy;
	exgcd(a, MOD, invx, invy);
	return (invx % MOD + MOD) % MOD;
}

int N, A[Max_N], C[Max_N], Fac[Max_N], Inv[Max_N], Ans, Cnt, inv2, Old[Max_N], Newa[Max_N], Newb[Max_N], Newm[Max_N], Pre[Max_N], Lst0[Max_N];

inline int work(int l, int r)
{
	return l > r ? 1 : (1 <= l && r <= N ? Mult(Fac[r], Inv[l - 1]) : 0);
}

int BIT[Max_N];

constexpr int lowbit(int x)
{
	return x & -x;
}

void insert(int i, int v)
{
	while (i)
		BIT[i] = Add(BIT[i], v), i -= lowbit(i);
}

int query(int i)
{
	int Ret(0);
	while (i <= N)
		Ret = Add(Ret, BIT[i]), i += lowbit(i);
	return Ret;
}

int main()
{
	inv2 = inverse(2);
	gi(N);
	for (int i = 1;i <= N;++i)
		gi(A[i]), ++C[A[i]];
	for (int x = 1;x <= N;++x)
		C[x] += C[x - 1];
	Fac[0] = 1;
	for (int i = 1;i <= N;++i)
		Fac[i] = Mult(Fac[i - 1], i);
	Inv[N] = inverse(Fac[N]);
	for (int i = N - 1;i >= 0;--i)
		Inv[i] = Mult(Inv[i + 1], i + 1);
	Cnt = 1;
	Pre[0] = 1;
	for (int x = 1;x <= N;++x)
	{
		Old[x] = work(x - C[x] + 1, x - C[x - 1]);
		if (Old[x] == 0)
		{
			printf("0");
			return 0;
		}
		Cnt = Mult(Cnt, Old[x]);
		Newa[x] = Mult(inverse(Old[x]), work(x - C[x], x - C[x - 1]));
		Newb[x] = Mult(inverse(Old[x]), work(x - C[x] + 1, x - C[x - 1] - 1));
		Newm[x] = Mult(inverse(Old[x]), work(x - C[x], x - C[x - 1] - 1));
		Lst0[x] = (Newm[x] == 0 ? x : Lst0[x - 1]);
		Pre[x] = Mult(Pre[x - 1], Newm[x] == 0 ? 1 : Newm[x]);
	}
	for (int x = 1;x <= N;++x)
		Ans = Add(Ans, Mult(Mult(Mult((C[x] - C[x - 1]), (C[x] - C[x - 1]) - 1), inv2), Mult(Cnt, inv2)));
	for (int i = 1, j;i <= N;++i)
	{
		j = Lst0[A[i] - 1];
		if (j == 0)
			j = 1;
		Ans = Add(Ans, Mult(inv2, Mult(Cnt, Mult(Sub(query(j), query(A[i])), Mult(Newb[A[i]], Pre[A[i] - 1])))));
		insert(A[i], Mult(Newa[A[i]], inverse(Pre[A[i]])));
	}
	memset(BIT, 0, sizeof(BIT));
	for (int i = N, j;i >= 1;--i)
	{
		j = Lst0[A[i] - 1];
		if (j == 0)
			j = 1;
		Ans = Sub(Ans, Mult(inv2, Mult(Cnt, Mult(Sub(query(j), query(A[i])), Mult(Newb[A[i]], Pre[A[i] - 1])))));
		insert(A[i], Mult(Newa[A[i]], inverse(Pre[A[i]])));
	}
	memset(BIT, 0, sizeof(BIT));
	for (int i = N, j;i >= 1;--i)
	{
		Ans = Add(Ans, Mult(Cnt, Sub(query(1), query(A[i]))));
		insert(A[i], 1);
	}
	print(Ans);
	return 0;
}
```