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>
#define ll long long
#define pii pair<int,int>
#define vec vector
using namespace std;

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	#ifdef LOCAL
//		freopen("in1.txt", "r", stdin);
//		freopen("in1.txt", "w", stdout);
	#endif
	ll n, q, k;
	cin >> n >> q >> k;
	vec<ll> a(n), pref(n);
	for (ll &x : a) cin >> x;
	for (int i=1; i<n-1; i++) pref[i] = pref[i-1] + a[i]-a[i-1]-1 + a[i+1]-a[i]-1;
	ll l, r, res;
	while (q--) {
		cin >> l >> r; --l; --r;
		if (l == r) cout << k-1 << '\n';
		else {
			res = max(0ll, pref[r-1] - pref[l]);
			res += a[l+1] - 2;
			res += k - a[r-1] - 1;
			cout << res << '\n';
		}
	}
}