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].
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define pb push_back

void solve()
{
    int n, q, k, l, r;
    cin >> n >> q >> k;
    vector<int> arr(n + 2);
    vector<ll> pre(n + 2);
    for (int i = 1; i <= n; i++)
        cin >> arr[i];
    arr[0] = 1;
    arr[n + 1] = k;
    pre[1] = arr[2] - 2;
    for (int i = 2; i < n; i++)
        pre[i] = pre[i - 1] + arr[i + 1] - arr[i - 1] - 2;
    pre[n] = k - arr[n - 1] - 1;
    // for (int i = 1; i <= n; i++)
    //     cout << pre[i] << " ";
    // cout << "\n";
    for (int i = 0; i < q; i++)
    {
        cin >> l >> r;
        if (l == r)
        {
            cout << k - 1 << "\n";
            continue;
        }
        if (r - l == 1)
        {
            cout << arr[r] - 2 + k - arr[l] - 1 << "\n";
            continue;
        }
        ll cur = pre[r - 1] - pre[l];
        cur += arr[l + 1] - 2 + k - arr[r - 1] - 1;
        cout << cur << "\n";
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
}