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 ll;
typedef long double ld;
typedef pair<ll, ll> pi;
typedef pair<double, double> pd;
typedef vector<ll> vi;
typedef vector<vector<ll>> vvi;
typedef vector<vector<pi>> vvpi;
typedef vector<pi> vpi;

ll INF = 2e18;
double eps = 1e-12;

#define forn(i, e) for (ll i = 0; i < e; i++)
#define forsn(i, s, e) for (ll i = s; i < e; i++)
#define rforn(i, s) for (ll i = s; i >= 0; i--)
#define rforsn(i, s, e) for (ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout << #x << " = " << x << ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())

void fast_cin()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

void read(vi &x)
{
    for (ll i = 0; i < sz(x); i++)
    {
        cin >> x[i];
    }
    return;
}

void read(vvi &x)
{
    for (ll i = 0; i < sz(x); i++)
    {
        for (ll j = 0; j < sz(x[i]); j++)
        {
            cin >> x[i][j];
        }
    }
    return;
}

void solve()
{
    ll n, q, k;
    cin >> n >> q >> k;
    vi a(n);
    read(a);
    vi pre(n), post(n), mid(n);
    forn(i, n)
    {
        if (i == 0)
        {
            pre[i] = k - 1;
            post[i] = a[i + 1] - 2;
            continue;
        }
        else if (i == n - 1)
        {
            pre[i] = k - a[i - 1] - 1;
            post[i] = k - 1;
            continue;
        }
        pre[i] = k - a[i - 1] - 1;
        post[i] = a[i + 1] - 2;
        mid[i] = a[i + 1] - a[i - 1] - 2;
    }
    for (ll i = 2; i < n - 1; i++)
    {
        mid[i] += mid[i - 1];
    }
    // forn(i,n)
    // {
    //     dbg(pre[i]);
    //     dbg(post[i]);
    //     dbg(mid[i]);
    // }
    ll l, r;
    while (q--)
    {

        cin >> l >> r;

        if (l == r)
        {
            cout << k - 1 << ln;
            continue;
        }

        ll tmp = (l + 1 < r) ? (mid[r - 2] - mid[l - 1]) : 0;
        ll ans = post[l - 1] + pre[r - 1] + tmp;
        cout << ans << ln;
    }
    return;
}

int main()
{
    // #ifndef ONLINE_JUDGE
    //     freopen("input.txt", "r", stdin);
    //     freopen("output.txt", "w", stdout);
    // #endif
    fast_cin();
    ll t = 1;
    // cin >> t;
    for (int it = 1; it <= t; it++)
    {
        // cout << "Case #" << it + 1 << ": ";
        solve();
    }
    return 0;
}