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 IOS ios_base::sync_with_stdio(0);cin.tie(0)
#define infinite 9123456789
#define PRECISION cout.precision(3); cout.setf(ios::fixed);
#define db cout << "Debug" << "\n";
#define ler_entrada freopen("test_input.txt", "r", stdin);

int n, q, k, a[100010];

void solve(){
	cin >> n >> q >> k;
	
	for(int i=1; i<=n; i++) cin >> a[i];
	vector<int> pref(n+10);
	
	pref[1] = a[2] - 2;
	
	for(int i=2; i<n; i++) pref[i] = pref[i-1] + a[i+1]-a[i-1]-2;
	
	pref[n] = k - a[n-1] - 1 + pref[n-1];
	
	for(int i=1; i<=q; i++){
		int l, r; cin >> l >> r;
		
		int quant = r - l + 1, dentro = a[r] - a[l] + 1 - quant;
		
		cout << k - quant + dentro << "\n";
		continue;
		
		if(l == r){
			cout << k-1 << "\n";
			continue;
		}
		if(l+1 == r){
			cout << k-2 << "\n";
			continue;
		}
		
		int resp = 0;
		resp += pref[r-1]-pref[l+1] + k - 1;
		cout << resp << "\n";
	}
}

signed main(){
	IOS;
	solve();
}