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 <bits/stdc++.h>
#define int long long 
using namespace std;

#define MAX 1000005

void solve(){

	int n,q,k; cin>>n>>q>>k;
	vector<int> a;
	int nn = n;
	while(nn--){
		int val; cin>>val;
		a.push_back(val);
	}
	vector<vector<int>> qq;
	for(int i=0;i<q;i++){
		int u,v; cin>>u>>v;
		qq.push_back({u,v});
	}
	vector<int> pre({0});
	
	for(int i=0;i<n;i++){
		if(i == 0){
			int val = a[i+1] - 0 - 2;
			pre.push_back(val);
		} else if(i==n-1){
			int val = k - a[i-1] - 1;
			pre.push_back(val + pre.back());
		} else {
			int val = a[i+1] - a[i-1] - 2;
			pre.push_back(val + pre.back());
		} 
	}

	// for(int i=0;i<=n;i++) cout<<pre[i]<<" ";
	// cout<<endl;

	for(auto el : qq){

		if(el[0] == n || el[1] == 1){
			cout<<k - 1<<endl;
			continue;
		}
		int val = pre[el[1]-1] - pre[el[0]] + a[el[0]] - 2 + k - a[el[1]-2] - 1;
		cout<<val<<endl;
	}

}

int32_t main(){

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	int t;
	// cin>>t;
	t = 1;

	while(t--){
		solve();
	}
}
