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].
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include <bits/stdc++.h>
#include <unordered_set>

int main() {
#ifdef DEBUG
	freopen("in.txt", "r", stdin);
	//    freopen("output.txt", "w", stdout);
#endif
	std::ios_base::sync_with_stdio(0); std::cin.tie(0); std::cout.tie(0);

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

	std::vector<int> v(n);
	for (auto& x : v) std::cin >> x;

	std::vector<long long> acums(n);
	long long acum = 0;
	for (int i = 0; i < n; i++) {
		int l = 0;
		int r = k + 1;
		if (i - 1 >= 0) l = v[i - 1];
		if (i + 1 < n) r = v[i + 1];
		acum += r - 1ll - l - 1ll;
		acums[i] = acum;
	}

	for (int i = 0; i < q; i++) {
		int li, ri;
		std::cin >> li >> ri;
		li--; ri--;

		long long ans;
		if (li == ri)
			ans = k - 1;
		else 
			ans = v[li + 1] - 2ll + (acums[ri - 1] - acums[li]) + (k - v[ri - 1] - 1);

		std::cout << ans << '\n';
	}
	return 0;
}