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<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#include<vector>
using namespace std;

typedef long long ll;
const int MAXN = 1e5 + 5;
const int inf = 0x3f3f3f3f;

int a[MAXN];
ll dp[MAXN];

int main()
{
	/*int t;
	cin >> t;
	while (t--) {*/
		int n, q, k;
		scanf("%d%d%d", &n, &q, &k);
		a[0] = 0, a[n + 1] = k + 1;
		dp[0] = 0;
		for (int i = 1;i <= n;i++) {
			scanf("%d", &a[i]);
		}
		for (int i = 1;i <= n;i++) {
			dp[i] = dp[i - 1] + (a[i + 1] - a[i - 1] - 1);
		}
		while (q--) {
			int l, r;
			scanf("%d%d", &l, &r);
			ll ans = dp[r] - dp[l - 1] - (r - l + 1);
			ans += a[l - 1];
			ans += k - a[r + 1] + 1;
			printf("%lld\n", ans);
		}
	/*}*/
	return 0;
}