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>
using namespace std;
#define lop(i,n) for(ll i=0;i<n;i++)
#define lop1(i,n) for(ll i=1;i<=n;i++)
#define lopr(i,n) for(ll i=n-1;i>=0;i--)
#define ll long long int
#define pb push_back
#define all(v) v.begin(),v.end()
#define IOS  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define endl '\n'
#define F first
#define S second	
#define mem(arr,val) memset(arr,val,sizeof(arr))
#define pii pair<int,int>
#define pll pair<ll,ll>
#define LCM(a,b) (a*b)/__gcd(a,b)
#define mii map<int,int>
#define mll map<ll,ll>
#define ub upper_bound
#define lb lower_bound
#define sz(x) (ll)x.size()
#define ld long double
#define pcnt(x) __builtin_popcountll(x)
const long long I1=1e9;
const long long I2=1e18;
const int32_t M1=1e9+7;
const int32_t M2=998244353;
template<typename T,typename T1>T maxn(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T minn(T &a,T1 b){if(b<a)a=b;return a;}

void solve()
{
  //code begins from here//
    int n,q,k;
    cin>>n>>q>>k;
    ll a[n+2];
    lop1(i,n){
        cin>>a[i];
    }
    a[0] = 0;
    a[n+1] = k+1;
    ll pre[n+2];
    lop1(i,n){
        pre[i] = (a[i]-a[i-1]-1)+(a[i+1]-a[i]-1);
        pre[i]+=pre[i-1];
    }
    while(q--){
        int l,r;
        cin>>l>>r;
        if (l==r){
            cout << k-1<<endl;
            continue;
        }
        ll cnt = 0;
        cnt += a[l+1]-2;
        cnt += k - a[r-1] -1;
        l++; r--;
        if (l<=r){
            cnt += pre[r]-pre[l-1];
        }
        cout << cnt<<endl;
    }
}

signed main()
{
    IOS;
    #ifdef MODULO
        initialize();
    #endif
    #ifdef SIEVE
        sieve();
    #endif
    int testcase=1;
   // cin>>testcase;
    while(testcase--) solve();
    return 0;
}