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 <vector>
#include <map>
#include <algorithm>
#include <set>
#include <math.h>
#include <queue>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <random>
using namespace std;
mt19937 ran(time(0));
int mod = 1e9 + 7;
long long Bpow(long long a, long long s) {
	if (s == 0) {
		return 1;
	}
	long long u = Bpow(a, s / 2);
	if (s & 1)
		return u * u % mod * a % mod;
	else
		return u * u % mod;
}
int main()
{
	int n, q, k;
	cin >> n >> q >> k;
	vector<int>a(n+2);
	a[n + 1] = k + 1;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	vector<int>sum(n + 2);
	for (int i = 1; i <= n; i++)
	{
		sum[i] = sum[i - 1] + a[i + 1] - a[i - 1] - 2;
	}
	while (q--) {
		int l, r;
		cin >> l >> r;
		int ans = sum[r] - sum[l - 1]+a[l-1]+k+1-a[r+1];
		cout << ans << "\n";
	}
}