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<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<functional>
#include<queue>
#include<map>
#include<cstring>
 
#define rep(i,x,n) for(int x = i; x <= n; ++x)
#define FOR0(x, n) for(int x = 0; x < n; ++x)
#define FOR1(x, n) for(int x = 1; i <= n; ++x)
#define vt vector
#define endl "\n"
#define all(c) (c).begin(), (c).end()
#define msl(arr) memset(arr,0x3f,sizeof(arr))
#define mss(arr) memset(arr,0x7f,sizeof(arr))
#define msz(arr) memset(arr,0,sizeof(arr))
#define INF 987654321
#define sz(x) (int)(x).size()

#define MAX 100010
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;

ll arr[MAX];
ll dp[MAX];
int T,n,q,k;
void solve(){
	arr[0] = 0;
	arr[n+1] = k;
	for(int i = 1; i <= n; i++){
		cin >> arr[i];	
	}
	for(int i = 1; i <= n; i++){
		dp[i] = arr[i] - arr[i-1] -1 + dp[i-1];
	}
	
	for(int i = 1; i <= q; i++){
		int s, e;
		cin >> s >> e;
		cout << (dp[e] - dp[s]) * 2 + arr[s] - 1 + k - arr[e] << "\n";
	}
	
	
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cin >> n >> q >> k;
	solve();
	cout << endl;

	return 0;
}