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].
// #pragma GCC optimize("Ofast")
#pragma GCC target ("sse4")
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
using namespace std;

const int N = 2e5 + 10, M = N * 2, INT = 0x3f3f3f3f, mod = 1e9 + 7;

//#define int long long
#define pi 3.1415926
#define x first
#define y second
typedef long long ll;
typedef pair<int,int> pii;
 //ios::sync_with_stdio(false);
ll gcd(ll a,ll b) {return b ? gcd(b, a % b) : a;}
// int gcd(int a,int b) {return b ? gcd(b, a % b) : a;}
ll qmi(ll a,ll b) {ll res = 1 % mod;while (b){if (b & 1) res = (res * a) % mod;a = a * a % mod;b >>= 1;}return res;}

int t, n, m, k;
int l, r;
int a[N], b[N];


void solve()
{
	cin >> n >> m >> k;

	for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
	a[n + 1] = k + 1;

	for (int i = 1; i <= n; i ++ )
	{
		b[i] = a[i + 1] - a[i - 1] - 2 + b[i - 1];

	}
	puts("");
	for (int i = 1; i <= m; i ++ )
	{
		scanf("%d%d", &l, &r);
		cout << b[r] - b[l - 1] + a[l - 1] + k + 1 - a[r + 1] << endl;
	}
}

signed main()
{
    // cin >> t;while (t -- ) 
    solve();
}