#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#include <bits/stdc++.h>
using namespace std;

#define fore(i, l, r) for (auto i = (l) - ((l) > (r)); i != (r) - ((l) > (r)); i += 1 - 2 * ((l) > (r)))
#define ceil(n, k) (((n) + ((k) - 1)) / (k))
#define mod(n, k) (((n) + (k)) % (k))
#define all(x) begin(x), end(x)
#define sz(x) int(x.size())
#define pb push_back
#define f first
#define s second

using ld = long double;
using lli = long long;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;

#ifdef LOCAL
  #include "../debug.h"
  const int N = 10;
#else
  #define debug(...)
  const int N = 1e6 + 5;
#endif

const int inf = numeric_limits<decltype(inf)>::max() / 2;

lli n, q, k;
lli a[N];
lli ps[N];

void testCase() {
  cin >> n >> q >> k;
  int i = 0;

  fore (j, 0, n) {
    cin >> a[j];
    if (a[j] <= k)
      i = j;
  }

  fore (i, 0, n) {
    if (i == 0) 
      ps[i] = a[i] - 1;
    if (i > 0)
      ps[i] += a[i] - a[i - 1] - 1;
    if (i < n - 1)
      ps[i] += a[i + 1] - a[i] - 1;
  }

  debug(ps);
  fore (i, 1, n)
    ps[i] += ps[i - 1];

  debug(ps);

  auto sum = [&] (int l, int r) {
    return ps[r] - (l == 0 ? 0 : ps[l - 1]);
  };

  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    r = min(r, i);
    debug(l, r, k, i);
    int ans = sum(l, r) + k - a[r];
    debug(ans);
    if (r < n - 1)
      ans -= a[r + 1] - a[r] - 1; 
    if (l > 0) {
      ans += a[l] - 1;
      ans -= a[l] - a[l - 1] - 1;
    }
    cout << ans << '\n';
  }
}

int main() {
  cin.tie(0)->sync_with_stdio(0), cout.tie(0);

  int tc = 1;
  // cin >> tc;

  while (tc--)
    testCase(); 

  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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].

### Response

```cpp
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
#include <bits/stdc++.h>
using namespace std;

#define fore(i, l, r) for (auto i = (l) - ((l) > (r)); i != (r) - ((l) > (r)); i += 1 - 2 * ((l) > (r)))
#define ceil(n, k) (((n) + ((k) - 1)) / (k))
#define mod(n, k) (((n) + (k)) % (k))
#define all(x) begin(x), end(x)
#define sz(x) int(x.size())
#define pb push_back
#define f first
#define s second

using ld = long double;
using lli = long long;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;

#ifdef LOCAL
  #include "../debug.h"
  const int N = 10;
#else
  #define debug(...)
  const int N = 1e6 + 5;
#endif

const int inf = numeric_limits<decltype(inf)>::max() / 2;

lli n, q, k;
lli a[N];
lli ps[N];

void testCase() {
  cin >> n >> q >> k;
  int i = 0;

  fore (j, 0, n) {
    cin >> a[j];
    if (a[j] <= k)
      i = j;
  }

  fore (i, 0, n) {
    if (i == 0) 
      ps[i] = a[i] - 1;
    if (i > 0)
      ps[i] += a[i] - a[i - 1] - 1;
    if (i < n - 1)
      ps[i] += a[i + 1] - a[i] - 1;
  }

  debug(ps);
  fore (i, 1, n)
    ps[i] += ps[i - 1];

  debug(ps);

  auto sum = [&] (int l, int r) {
    return ps[r] - (l == 0 ? 0 : ps[l - 1]);
  };

  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    r = min(r, i);
    debug(l, r, k, i);
    int ans = sum(l, r) + k - a[r];
    debug(ans);
    if (r < n - 1)
      ans -= a[r + 1] - a[r] - 1; 
    if (l > 0) {
      ans += a[l] - 1;
      ans -= a[l] - a[l - 1] - 1;
    }
    cout << ans << '\n';
  }
}

int main() {
  cin.tie(0)->sync_with_stdio(0), cout.tie(0);

  int tc = 1;
  // cin >> tc;

  while (tc--)
    testCase(); 

  return 0;
}

```