#include <iostream>
#include <algorithm>


#define for_n(i, n) for (int i = 0; i < (int) n; i++)

using namespace std;

const int MAX_N_SIZE = 1e5+ 5;
int digits[MAX_N_SIZE];

unsigned long long int solve(int k, int l, int r) {

    return k + digits[r - 1] - digits[l - 1] + 1 - 2 * (r - l + 1);
}

int main() {

    int n, q, k, l, r;
    cin >> n >> q >> k;

    for_n(i, n) cin >> digits[i];

    while (q--) {
        cin >> l >> r;
        cout << solve(k, l , r) << endl;
    }

    return 0;
}


### Prompt

Develop a solution in cpp to the problem described below:

Given a positive integer k, two arrays are called k-similar if:

  * they are strictly increasing; 
  * they have the same length; 
  * all their elements are positive integers between 1 and k (inclusive); 
  * they differ in exactly one position. 



You are given an integer k, a strictly increasing array a and q queries. For each query, you are given two integers l_i ≤ r_i. Your task is to find how many arrays b exist, such that b is k-similar to array [a_{l_i},a_{l_i+1}…,a_{r_i}]. 

Input

The first line contains three integers n, q and k (1≤ n, q ≤ 10^5, n≤ k ≤ 10^9) — the length of array a, the number of queries and number k.

The second line contains n integers a_1, a_2, …,a_n (1 ≤ a_i ≤ k). This array is strictly increasing — a_1 < a_2 < … < a_n.

Each of the following q lines contains two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

Print q lines. The i-th of them should contain the answer to the i-th query.

Examples

Input


4 2 5
1 2 4 5
2 3
3 4


Output


4
3


Input


6 5 10
2 4 6 7 8 9
1 4
1 2
3 5
1 6
5 5


Output


8
9
7
6
9

Note

In the first example:

In the first query there are 4 arrays that are 5-similar to [2,4]: [1,4],[3,4],[2,3],[2,5].

In the second query there are 3 arrays that are 5-similar to [4,5]: [1,5],[2,5],[3,5].

### Response

```cpp
#include <iostream>
#include <algorithm>


#define for_n(i, n) for (int i = 0; i < (int) n; i++)

using namespace std;

const int MAX_N_SIZE = 1e5+ 5;
int digits[MAX_N_SIZE];

unsigned long long int solve(int k, int l, int r) {

    return k + digits[r - 1] - digits[l - 1] + 1 - 2 * (r - l + 1);
}

int main() {

    int n, q, k, l, r;
    cin >> n >> q >> k;

    for_n(i, n) cin >> digits[i];

    while (q--) {
        cin >> l >> r;
        cout << solve(k, l , r) << endl;
    }

    return 0;
}


```