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 int long long

void Solve() {
    int n, q, k;
    cin >> n >> q >> k;
    vector<int> a(n + 2);
    int left[n + 1], right[n + 1], pref[n + 1];
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    a[0] = 0;
    a[n + 1] = k + 1;
    pref[0] = 0;
    for(int i = 1; i <= n; i++) {
        left[i] = a[i - 1] + 1;
        right[i] = a[i + 1] - 1;
        pref[i] = pref[i - 1] + right[i] - left[i];
    }
    for(int i = 1; i <= q; i++) {
        int l, r;
        cin >> l >> r;
        if(l == r) {
            cout << k - 1 << "\n";
            continue;
        }
        else {
            int ans = pref[r - 1] - pref[l];
            int up = a[l + 1];
            ans += up - 2;
            up = a[r - 1];
            ans += k - up - 1;
            cout << ans << "\n";
        }
    }
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    //cin >> t;
    while(t--) {
        Solve();
    }
}
