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;

typedef long long ll;
#define int ll
#define endl '\n'
#define yes cout << "YES" << endl
#define no cout << "NO" << endl

#ifdef ROHIT_SYS
#include "headers/debug.hpp"
#endif

 
int32_t main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
 
	#ifdef ROHIT_SYS
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	#endif

	int n, q, k;
	cin >> n >> q >> k;

	vector<int> v(n), pref(n, 0);
	for(int i = 0; i < n; i++) cin >> v[i];
	for(int i = 1; i < n-1; i++) pref[i] = v[i+1]-v[i-1]-2;
	for(int i = 1; i < n; i++) pref[i] += pref[i-1];

	int x, y;
	while(q--) {
		cin >> x >> y;
		x--, y--;

		int res;
		if(x == y) {
			cout << k-1 << endl;
			continue;
		} else res = (v[x+1]-2) + (k-v[y-1]-1);

		res += pref[y-1]-pref[x];

		cout << res << endl;
	}
 
	return 0;
}

