#include <bits/stdc++.h>
using namespace std;
const long long mod = (long long)1e9 + 7;
long long fastpow(long long b, long long e) {
  if (e == 0) {
    return 1;
  }
  if (e == 1) {
    return b;
  }
  long long a = fastpow(b, e / 2) % mod;
  a = (a * a) % mod;
  a = (a * fastpow(b, e % 2)) % mod;
  return a % mod;
}
int main(int argc, char const *argv[]) {
  std::ios_base::sync_with_stdio(false);
  long long n, q, m;
  cin >> n >> q >> m;
  vector<long long> arr(n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<pair<long long, pair<long long, long long> > > que(q);
  for (long long i = 0; i < q; i++) {
    cin >> que[i].first >> que[i].second.first >> que[i].second.second;
    que[i].second.first--;
    que[i].second.second--;
  }
  vector<long long> spec(m, 0);
  for (long long i = 0; i < m; i++) {
    cin >> spec[i];
    spec[i]--;
  }
  for (long long i = que.size() - 1; i >= 0; i--) {
    for (long long mi = 0; mi < spec.size(); mi++) {
      if (spec[mi] >= que[i].second.first && spec[mi] <= que[i].second.second) {
        if (que[i].first == 1) {
          if (spec[mi] == que[i].second.first) {
            spec[mi] = que[i].second.second;
          } else {
            spec[mi] = spec[mi] - 1;
          }
        } else {
          spec[mi] = que[i].second.second + que[i].second.first - spec[mi];
        }
      }
    }
  }
  for (long long i = 0; i < spec.size(); i++) {
    cout << arr[spec[i]] << ' ';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given an array a of size n, and q queries to it. There are queries of two types: 

  * 1 li ri — perform a cyclic shift of the segment [li, ri] to the right. That is, for every x such that li ≤ x < ri new value of ax + 1 becomes equal to old value of ax, and new value of ali becomes equal to old value of ari; 
  * 2 li ri — reverse the segment [li, ri]. 



There are m important indices in the array b1, b2, ..., bm. For each i such that 1 ≤ i ≤ m you have to output the number that will have index bi in the array after all queries are performed.

Input

The first line contains three integer numbers n, q and m (1 ≤ n, q ≤ 2·105, 1 ≤ m ≤ 100). 

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109). 

Then q lines follow. i-th of them contains three integer numbers ti, li, ri, where ti is the type of i-th query, and [li, ri] is the segment where this query is performed (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n). 

The last line contains m integer numbers b1, b2, ..., bm (1 ≤ bi ≤ n) — important indices of the array. 

Output

Print m numbers, i-th of which is equal to the number at index bi after all queries are done.

Example

Input

6 3 5
1 2 3 4 5 6
2 1 3
2 3 6
1 1 6
2 2 1 5 3


Output

3 3 1 5 2 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = (long long)1e9 + 7;
long long fastpow(long long b, long long e) {
  if (e == 0) {
    return 1;
  }
  if (e == 1) {
    return b;
  }
  long long a = fastpow(b, e / 2) % mod;
  a = (a * a) % mod;
  a = (a * fastpow(b, e % 2)) % mod;
  return a % mod;
}
int main(int argc, char const *argv[]) {
  std::ios_base::sync_with_stdio(false);
  long long n, q, m;
  cin >> n >> q >> m;
  vector<long long> arr(n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<pair<long long, pair<long long, long long> > > que(q);
  for (long long i = 0; i < q; i++) {
    cin >> que[i].first >> que[i].second.first >> que[i].second.second;
    que[i].second.first--;
    que[i].second.second--;
  }
  vector<long long> spec(m, 0);
  for (long long i = 0; i < m; i++) {
    cin >> spec[i];
    spec[i]--;
  }
  for (long long i = que.size() - 1; i >= 0; i--) {
    for (long long mi = 0; mi < spec.size(); mi++) {
      if (spec[mi] >= que[i].second.first && spec[mi] <= que[i].second.second) {
        if (que[i].first == 1) {
          if (spec[mi] == que[i].second.first) {
            spec[mi] = que[i].second.second;
          } else {
            spec[mi] = spec[mi] - 1;
          }
        } else {
          spec[mi] = que[i].second.second + que[i].second.first - spec[mi];
        }
      }
    }
  }
  for (long long i = 0; i < spec.size(); i++) {
    cout << arr[spec[i]] << ' ';
  }
  return 0;
}

```