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 ll long long

void solve() {
    ll n, q, k; cin >> n >> q >> k;
    vector<ll> A(n);
    for (ll &a : A)
        cin >> a;
    vector<ll> P(n);
    if (n > 1)
        P[0] = A[1] - 2;
    for (int i = 1; i < n - 1; i++) {
        P[i] = A[i + 1] - A[i - 1] - 2;
        if (i)
            P[i] += P[i - 1];
    }
    P[n - 1] = k - A[n - 1];
    if (n > 1)
        P[n - 1] += P[n - 2];
    // for (int i = 0; i < n; i++)
    //     cout << P[i] << " ";
    // cout << endl;
    auto f = [&](int l, int r) {
        if (l == r) {
            return k - 1;
        }
        ll ans = P[r - 1] - P[l];
        ans += A[l + 1] - 2;
        ans += k - A[r - 1] - 1;
        // cout << l << " " << r << " " << P[r - 1] - P[l] << " " << A[l + 1] - 2 << " " << k - A[r - 1] - 1 << endl;
        return ans;
    };
    while (q--) {
        int l, r; cin >> l >> r;
        l--, r--;
        cout << f(l, r) << '\n';
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // int t; cin >> t;
    // while (t--)
        solve();
    return 0;
}