#include <cmath>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>

using namespace std;

typedef long long ll;

mt19937 rnd(228);

const int M = 1e9 + 7;

inline int add(int a, int b)
{
    a += b;
    if (a >= M) a -= M;
    if (a < 0) a += M;
    return a;
}

inline int mul(int a, int b)
{
    return (a * (ll) b) % M;
}

const int N = 2e5 + 7;

int t[4 * N];
int upd_add[4 * N];
int upd_mul[4 * N];
int cnt[4 * N];

void upd(int v, int k, int b)
{
    t[v] = add(mul(t[v], k), mul(b, cnt[v]));
    upd_add[v] = add(mul(upd_add[v], k), b);
    upd_mul[v] = mul(upd_mul[v], k);
}

void push(int v)
{
    upd(v * 2 + 1, upd_mul[v], upd_add[v]);
    upd(v * 2 + 2, upd_mul[v], upd_add[v]);
    upd_mul[v] = 1, upd_add[v] = 0;
}

void upd(int v, int l, int r, int tl, int tr, int k, int b)
{
    if (tl >= r || tr <= l)
    {
        return;
    }
    if (tl >= l && tr <= r)
    {
        upd(v, k, b);
    }
    else
    {
        int tm = (tl + tr) / 2;
        push(v);
        upd(v * 2 + 1, l, r, tl, tm, k, b);
        upd(v * 2 + 2, l, r, tm, tr, k, b);
        t[v] = add(t[v * 2 + 1], t[v * 2 + 2]);
    }
}

int get(int v, int l, int r, int tl, int tr)
{
    if (tl >= r || tr <= l)
    {
        return 0;
    }
    if (tl >= l && tr <= r)
    {
        return t[v];
    }
    else
    {
        int tm = (tl + tr) / 2;
        push(v);
        return add(get(v * 2 + 1, l, r, tl, tm), get(v * 2 + 2, l, r, tm, tr));
    }
}

void build(int v, int l, int r)
{
    upd_mul[v] = 1;
    upd_add[v] = t[v] = 0;
    cnt[v] = r - l;
    if (r - l != 1)
    {
        int m = (l + r) / 2;
        build(v * 2 + 1, l, m);
        build(v * 2 + 2, m, r);
    }
}

int fenw[N];

void inc(int x, int y)
{
    for (; x < N; x = (x | (x + 1)))
    {
        fenw[x] += y;
    }
}

int get(int x)
{
    int ans = 0;
    for (; x >= 0; x = (x & (x + 1)) - 1)
    {
        ans += fenw[x];
    }
    return ans;
}

int suf(int x)
{
    return get(N - 1) - get(x);
}

int main()
{
#ifdef ONPC
    freopen("a.in", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    build(0, 0, n + 1);
    vector <int> a(n);
    vector <int> cnt(n + 1);
    vector <vector <int> > on(n + 1);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        on[n - a[i] + 1].push_back(i + 1);
        cnt[a[i]]++;
    }
    for (int i = n - 1; i >= 0; i--)
    {
        cnt[i] += cnt[i + 1];
    }
    for (int i = n; i >= 1; i--)
    {
        if (cnt[i] <= n - i)
        {
            cout << 0 << '\n';
            return 0;
        }
    }
    reverse(cnt.begin() + 1, cnt.end());
    vector <int> dp(n + 1);
    vector <int> ct(n + 1);
    dp[0] = 1;
    cnt[0] = 0;
    int rev_2 = (M + 1) / 2;
    int sum = 0;
    for (int i = 1; i <= n; i++) inc(i, 1);
    for (int i = 1; i <= n; i++)
    {
        for (int go : on[i])
        {
            sum = add(sum, -get(0, 0, go, 0, n + 1));
            inc(go, -1);
        }
        for (int go : on[i])
        {
            upd(0, go, go + 1, 0, n + 1, 1, dp[i - 1]);
            sum = add(sum, mul(dp[i - 1], suf(go)));
        }
        int x = cnt[i] - (i - 1);
        int x_sqr = mul(mul(x, (x - 1)), rev_2);
        ct[i] = add(add(mul(x_sqr, dp[i - 1]), mul(ct[i - 1], x)), sum);
        dp[i] = mul(dp[i - 1], x);
        sum = mul(sum, (x - 1));
        upd(0, 0, n + 1, 0, n + 1, x - 1, 0);
    }
    cout << ct[n] << '\n';
}

### Prompt

Generate 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 <cmath>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>

using namespace std;

typedef long long ll;

mt19937 rnd(228);

const int M = 1e9 + 7;

inline int add(int a, int b)
{
    a += b;
    if (a >= M) a -= M;
    if (a < 0) a += M;
    return a;
}

inline int mul(int a, int b)
{
    return (a * (ll) b) % M;
}

const int N = 2e5 + 7;

int t[4 * N];
int upd_add[4 * N];
int upd_mul[4 * N];
int cnt[4 * N];

void upd(int v, int k, int b)
{
    t[v] = add(mul(t[v], k), mul(b, cnt[v]));
    upd_add[v] = add(mul(upd_add[v], k), b);
    upd_mul[v] = mul(upd_mul[v], k);
}

void push(int v)
{
    upd(v * 2 + 1, upd_mul[v], upd_add[v]);
    upd(v * 2 + 2, upd_mul[v], upd_add[v]);
    upd_mul[v] = 1, upd_add[v] = 0;
}

void upd(int v, int l, int r, int tl, int tr, int k, int b)
{
    if (tl >= r || tr <= l)
    {
        return;
    }
    if (tl >= l && tr <= r)
    {
        upd(v, k, b);
    }
    else
    {
        int tm = (tl + tr) / 2;
        push(v);
        upd(v * 2 + 1, l, r, tl, tm, k, b);
        upd(v * 2 + 2, l, r, tm, tr, k, b);
        t[v] = add(t[v * 2 + 1], t[v * 2 + 2]);
    }
}

int get(int v, int l, int r, int tl, int tr)
{
    if (tl >= r || tr <= l)
    {
        return 0;
    }
    if (tl >= l && tr <= r)
    {
        return t[v];
    }
    else
    {
        int tm = (tl + tr) / 2;
        push(v);
        return add(get(v * 2 + 1, l, r, tl, tm), get(v * 2 + 2, l, r, tm, tr));
    }
}

void build(int v, int l, int r)
{
    upd_mul[v] = 1;
    upd_add[v] = t[v] = 0;
    cnt[v] = r - l;
    if (r - l != 1)
    {
        int m = (l + r) / 2;
        build(v * 2 + 1, l, m);
        build(v * 2 + 2, m, r);
    }
}

int fenw[N];

void inc(int x, int y)
{
    for (; x < N; x = (x | (x + 1)))
    {
        fenw[x] += y;
    }
}

int get(int x)
{
    int ans = 0;
    for (; x >= 0; x = (x & (x + 1)) - 1)
    {
        ans += fenw[x];
    }
    return ans;
}

int suf(int x)
{
    return get(N - 1) - get(x);
}

int main()
{
#ifdef ONPC
    freopen("a.in", "r", stdin);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    build(0, 0, n + 1);
    vector <int> a(n);
    vector <int> cnt(n + 1);
    vector <vector <int> > on(n + 1);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        on[n - a[i] + 1].push_back(i + 1);
        cnt[a[i]]++;
    }
    for (int i = n - 1; i >= 0; i--)
    {
        cnt[i] += cnt[i + 1];
    }
    for (int i = n; i >= 1; i--)
    {
        if (cnt[i] <= n - i)
        {
            cout << 0 << '\n';
            return 0;
        }
    }
    reverse(cnt.begin() + 1, cnt.end());
    vector <int> dp(n + 1);
    vector <int> ct(n + 1);
    dp[0] = 1;
    cnt[0] = 0;
    int rev_2 = (M + 1) / 2;
    int sum = 0;
    for (int i = 1; i <= n; i++) inc(i, 1);
    for (int i = 1; i <= n; i++)
    {
        for (int go : on[i])
        {
            sum = add(sum, -get(0, 0, go, 0, n + 1));
            inc(go, -1);
        }
        for (int go : on[i])
        {
            upd(0, go, go + 1, 0, n + 1, 1, dp[i - 1]);
            sum = add(sum, mul(dp[i - 1], suf(go)));
        }
        int x = cnt[i] - (i - 1);
        int x_sqr = mul(mul(x, (x - 1)), rev_2);
        ct[i] = add(add(mul(x_sqr, dp[i - 1]), mul(ct[i - 1], x)), sum);
        dp[i] = mul(dp[i - 1], x);
        sum = mul(sum, (x - 1));
        upd(0, 0, n + 1, 0, n + 1, x - 1, 0);
    }
    cout << ct[n] << '\n';
}

```