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
#define ld long double
#define ll long long
#define vll vector<int>
#define mset(a) memset(a,0,sizeof(a))
#define pb push_back
#define range(c) c.begin() , c.end()
#define forn(i,n) for(ll i=0; i<(int)n; ++i)
#define for1(i,n) for(ll i=1; i<(int)n; ++i)
#define endl '\n'
#define MOD 1000000007
#define pii pair<int,int>
#define test ll t;cin>>t;while(t--)
using namespace std;

vector<bool> sieve(ll n){
    vector<bool>prime(n+1,true);
    prime[0]=prime[1]=false;
    for(ll i=2; i<=n; ++i){
        if(prime[i]&&(i*i)<=n)
        for(ll j=i*i; j<=n; j+=i)prime[j]=false;
    }
    return prime;
}

ll power(ll a, ll b, ll m = MOD) {
    a %= m;
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}

//////////////////////////////////////////////////////////////

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

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

    int n,q,k;
    cin>>n>>q>>k;

    vll a(n);
    forn(i,n) cin>>a[i];

    while(q--) {
        int l,r;
        cin>>l>>r;
        
        int res = k + (a[r-1] - a[l-1] + 1) - 2*(r-l+1);
        cout<<res<<endl;
    }    

    return 0;
}
