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].
// Problem: B. Replace and Keep Sorted
// Contest: Codeforces - Codeforces Round #701 (Div. 2)
// URL: https://codeforces.com/contest/1485/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <cstdio>
#include <cctype>

template <typename Tp>
void read(Tp &res){
	static char ch; ch = getchar(), res = 0;
	static bool flag; flag = 0;
	while(!isdigit(ch)) flag = (ch == '-'), ch = getchar();
	while(isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
	if(flag) res = -res;
}

const int maxn = 1e5 + 19;

int n, q, k, a[maxn], s[maxn];

int main(){
	read(n), read(q), read(k);
	for(int i = 1; i <= n; ++i) read(a[i]);
	for(int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i] - a[i - 1] - 1;
	while(q--){
		int l, r;
		read(l), read(r);
		printf("%lld\n", (long long)(s[r] - s[l]) * 2 + a[l] - 1 + k - a[r]);
	}
}