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 int long long
#define INF 1000000000000000000

int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int tt=1;
	//cin >> tt;
	while(tt--)
	{
		int n,q,k;
		cin >> n >> q >> k;
		int a[n+2];
		a[0]=0;
		for(int i=1;i<=n;i++)
			cin >> a[i];
		a[n+1]=k+1;
		int pre[n+1];
		for(int i=1;i<=n;i++)
			pre[i]=a[i+1]-a[i-1]-2;
		for(int i=2;i<=n;i++)
			pre[i]=pre[i-1]+pre[i];
		for(int i=0;i<q;i++)
		{
			int l,r;
			cin >> l >> r;
			if(l==r)
			{
				cout << k-1 <<'\n';
			}
			else if(r-l==1)
			{
				cout << a[r]-2+k-a[l]-1 << '\n';
			}
			else
			{
				int ans=a[l+1]-2+k-a[r-1]-1;
				r--;
				ans+=pre[r]-pre[l];
				cout << ans << '\n';
			}
		}
	}
	return 0;
}