#include <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto &&i : v) scanf("%d", &i);
    ll ans = 0;
    for (int i = 0; i < n; ++i) {
        ans += v[i]-i-1;
    }
    cout << ans << "\n";
    return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

Gag

Segtree has $ N $ of "gags", each with a value of $ V_i $.

Segtree decided to publish all the gags in any order.

Here, the "joy" you get when you publish the $ i $ th gag to the $ j $ th is expressed as $ V_i --j $.

Find the maximum sum of the "joy" you can get.

input

Input is given from standard input in the following format.


$ N $
$ V_1 $ $ V_2 $ $ \ ldots $ $ V_N $


output

Please output the maximum value of the sum of "joy". However, the value does not always fit in a 32-bit integer.

Insert a line break at the end.

Constraint

* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq V_i \ leq 10 ^ 5 $
* All inputs are integers.



Input example 1


1
59549


Output example 1


59548


Input example 2


Five
2 1 8 5 7


Output example 2


8






Example

Input

1
59549


Output

59548

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto &&i : v) scanf("%d", &i);
    ll ans = 0;
    for (int i = 0; i < n; ++i) {
        ans += v[i]-i-1;
    }
    cout << ans << "\n";
    return 0;
}

```