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;
#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl '\n'

void scanArr(ll a[], ll n)
{
    for(ll i=0;i<n;i++)
        cin>>a[i];
}

int main()
{
    fast;

    ll n, q, k, l, r; cin>>n>>q>>k;
    ll a[n]; scanArr(a, n);

    if(n==1)
    {
        while(q--)
        {
            cin>> l >> r;

            cout << k-1 << endl;
        }
        return 0;
    }


    ll lessA[n];
    ll pre[n+1];

    lessA[0] = a[1]-2;
    lessA[n-1] = k - a[n-2] - 1;

    pre[0] = 0;
    pre[1] = lessA[0];



    for(ll i=1;i<n-1;i++)
    {
        lessA[i] = a[i+1] - a[i-1] - 2;
        pre[i+1] = lessA[i] + pre[i];
    }
    pre[n] = lessA[n-1] + pre[n-1];

   /* for(ll i=0;i<n;i++)
        cout<< a[i] << " ";
    cout<<endl;
    for(ll i=0;i<n;i++)
        cout << lessA[i] << " ";
    cout<<endl;

    for(ll i=1;i<=n;i++)
        cout<<pre[i] << " ";
    cout<<endl;*/


    while(q--)
    {
        cin>>l>>r;

        if(l==r)
            cout<< k - 1 << endl;
        else
        {
            ll ans = 0;
            ans += max(a[l] - 2, 0ll);
            ans += max( k - a[r-2] - 1 , 0ll);
            l++;
            r--;
            cout<< ans + pre[r] - pre[l-1] << endl;
        }


    }
    return 0;
}
