#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fr first
#define sc second
#define mod 998244353
#define len(x) x.size()
#define min3(a, b, c) min(a, min(b, c))
#define max3(a, b, c) max(a, max(b, c))
#define all(v) v.begin(), v.end()
#define alla(a,n) a, a + n
#define fast_io ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<vll> vvll;
typedef vector<string> vs;

int32_t main() {

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    fast_io;
    
        
        ll n, q, k; cin>>n>>q>>k;
        ll arr[n];
        for(ll i=0;i<n;i++){
            cin>>arr[i];
        }

        ll ans[n];
        memset(ans, 0, sizeof(ll)*n);
        ans[0]+=arr[0]-1;
        ans[n-1]+=k-arr[n-1];
        for(ll i=0;i<n;i++){

            if(i!=0 && arr[i]!=arr[i-1]){
                ans[i]+=arr[i]-arr[i-1]-1;
            }
            if(i!=n-1 && arr[i]!=arr[i+1]){
                ans[i]+=arr[i+1]-arr[i]-1;
            }
        }

        for(ll i=1;i<n;i++){
            ans[i]+=ans[i-1];
            // cout<<ans[i]<<" ";
        }
        for(ll j=0;j<q;j++){

            ll x, y; cin>>x>>y;
            ll finalAns= ans[y-1];
            if(x>=2){
                finalAns-=ans[x-2];
            }

            if(x!=1){
                finalAns+=arr[x-2];
            }

            if(y!=n){
                finalAns+=k-arr[y]+1;
            }
            cout<<finalAns<<endl;
        }


    
    return 0;

}
### Prompt

Your task is to create a cpp solution to the following problem:

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].

### Response

```cpp
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fr first
#define sc second
#define mod 998244353
#define len(x) x.size()
#define min3(a, b, c) min(a, min(b, c))
#define max3(a, b, c) max(a, max(b, c))
#define all(v) v.begin(), v.end()
#define alla(a,n) a, a + n
#define fast_io ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<vll> vvll;
typedef vector<string> vs;

int32_t main() {

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    fast_io;
    
        
        ll n, q, k; cin>>n>>q>>k;
        ll arr[n];
        for(ll i=0;i<n;i++){
            cin>>arr[i];
        }

        ll ans[n];
        memset(ans, 0, sizeof(ll)*n);
        ans[0]+=arr[0]-1;
        ans[n-1]+=k-arr[n-1];
        for(ll i=0;i<n;i++){

            if(i!=0 && arr[i]!=arr[i-1]){
                ans[i]+=arr[i]-arr[i-1]-1;
            }
            if(i!=n-1 && arr[i]!=arr[i+1]){
                ans[i]+=arr[i+1]-arr[i]-1;
            }
        }

        for(ll i=1;i<n;i++){
            ans[i]+=ans[i-1];
            // cout<<ans[i]<<" ";
        }
        for(ll j=0;j<q;j++){

            ll x, y; cin>>x>>y;
            ll finalAns= ans[y-1];
            if(x>=2){
                finalAns-=ans[x-2];
            }

            if(x!=1){
                finalAns+=arr[x-2];
            }

            if(y!=n){
                finalAns+=k-arr[y]+1;
            }
            cout<<finalAns<<endl;
        }


    
    return 0;

}
```