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
const ll INF = 2e18;
#define int long long
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define vi vector<int>
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size();
#define endl "\n"
#define fastio      \
    cin.tie(NULL);  \
    cout.tie(NULL); \
    ios_base::sync_with_stdio(false)
#define double long double
#define rep(i, x, n) for (int i = x; i < (n); i++)
#define td(i) i.begin(), i.end()
#define rtd(i) i.rbegin(), i.rend()

const double e = exp(1);
const double pi = acos(-1.0);

int gcd(int a, int b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }

long long binpows(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1)  // impar
            res = res * a;
        a = a * a;
        b >>= 1;  // /2
    }
    return res;
}
ll binpow(ll a, ll b, ll m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1) res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}
void solve() {
    int n, q, k;
    cin >> n >> q >> k;
    vector<int> v(n + 2), pre(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> v[i];
    }
    v[n + 1] = k;
    for (int i = 1; i <= n; i++) {
        pre[i] = v[i + 1] - v[i - 1] - 2;
        pre[i] += pre[i - 1];
    }
    while (q--) {
        int l, r;
        cin >> l >> r;
        int qt = v[l + 1] - 2;
        qt += k - v[r - 1] - 1;
        qt += pre[r - 1] - pre[l];
        cout << qt << endl;
    }
}
// revise as variaveis e a ordem nas funções
signed main() {
    // freopen("zeros.in", "r", stdin);
    // freopen("hidden.out", "w", stdout);
    fastio;
    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }
}
