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 int
#define watch(x) cout << (#x) << " is " << (x) << endl
#define MOD 1000000007




void solve(int T)
{

  int n,q,k;
  cin>> n >> q >> k;
  vector<int> arr(n+2);
  for(int i=1;i<=n;i++){
    cin>>arr[i];
  }
	arr[0] = 0;
	arr[n+1] = k+1;
	int H[n+1];
	H[0] = 0;
	for(int  i = 1; i <=n ; i++){
		H[i] = arr[i+1] - arr[i-1] - 2;
	}

	// for(int i = 0 ; i <= n;i++){
	// 	cout << H[i] <<" ";
	// }
	int presum[n+1];
	partial_sum(H,H+n+1,presum);
	// cout << endl;
	// for(int i = 0 ; i <= n;i++){
	// 	cout << presum[i] <<" ";
	// }
	// cout << endl;
	for(int  i = 0; i < q; i++ ){
		int l , r;
		cin >> l >> r;
		if(l==r){
			cout << k-1 << endl;
			continue;
		}
		int ans = presum[r] - presum[l-1];
		// watch(ans);
		ans = ans - H[l] + (arr[l+1]-2) - H[r] + (k - arr[r-1] -1);
		cout << ans << endl;
	}
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

#ifndef ONLINE_JUDGE
    freopen("inputf.in", "r", stdin);
    freopen("error.out", "w", stderr);
    freopen("outputf.out", "w", stdout);
#endif

    int T = 1;
    //cin >> T;
    for (int t=1;t<=T; t++) {
        solve(t);
    }

    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
    return 0;
}