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;
typedef long long int ll;
typedef pair<ll, ll> pll;
#define pb push_back
typedef long double ld;
#define fi first
#define se second
#define PI 3.14159265358979323846264338327950288419716939937510
#define mp make_pair
#define vvl vector<vector<ll>>
#define vll vector<ll>
#define vpll vector<pll>
#define vvpll vector<vpll>
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
*/

ll inf = 1e18;
ll p = 1e9 + 7;
ld eps = 1e-9;
ll power(ll x, ll y, ll p)
{
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
ll pwr(ll x, ll y)
{
    ll res = 1;
    x = x ;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) ;
        y = y >> 1;
        x = (x * x) ;
    }
    return res;
}
ll modInverse(ll n, ll p)
{
    return power(n, p - 2, p);
}
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);

}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll i, y, x, z, g, key,   h, n, m, a, b, c, j, k, d, q;
    ll t2, t3, t4, t1, t5, t6;
    string s, s2, s1;
    ll l, r;
    ll tc;
    //cin >> tc;
    tc = 1;
    ll x1, y1, x2, y2;
    while (tc--)
    {
        cin >> n >> q >> k;
        ll arr[n];
        vll pre(n, 0);
        for (i = 0; i < n; i++)
            cin >> arr[i];
        for (i = 1; i < (n - 1); i++)
        {
            pre[i] = pre[i - 1] + (arr[i + 1] - arr[i - 1] - 2);
        }
        while (q--)
        {
            ll l, r;
            cin >> l >> r;
            if (l == r)
                cout << k - 1 << '\n';
            else
            {
                ll ans = pre[r - 2] - pre[l - 1];
                l--;
                r--;

                ans += arr[l + 1] - 2;
                ans += (k - arr[r - 1] - 1);
                cout << ans << "\n";
            }
        }







    }



    return 0;
}