#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int ls[N << 7], rs[N << 7];
int mn[N << 7], lazy[N << 7];
int Log[N], a[N], f[N][20];
int cnt = 1, q, n, k;
void init() {
  Log[0] = -1;
  for (int i = 1; i <= n; i++) f[i][0] = a[i], Log[i] = Log[i >> 1] + 1;
  for (int j = 1; j <= 19; j++)
    for (int i = 1; i + (1 << (j - 1)) <= n; i++)
      f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
int RMQ(int x, int y) {
  int t = Log[y - x + 1];
  return min(f[x][t], f[y - (1 << t) + 1][t]);
}
int qry(int l, int r) {
  if (r - l >= n)
    return RMQ(1, n);
  else {
    l = (l - 1) % n + 1, r = (r - 1) % n + 1;
    if (l > r)
      return min(RMQ(1, r), RMQ(l, n));
    else
      return RMQ(l, r);
  }
}
void pushdown(int p, int l, int r) {
  int mid = (l + r) >> 1;
  if (!ls[p]) ls[p] = ++cnt;
  if (!rs[p]) rs[p] = ++cnt;
  if (~lazy[p]) {
    lazy[ls[p]] = lazy[rs[p]] = lazy[p];
    mn[ls[p]] = lazy[ls[p]] ? lazy[ls[p]] : qry(l, mid);
    mn[rs[p]] = lazy[rs[p]] ? lazy[rs[p]] : qry(mid + 1, r);
  }
}
void update(int x, int y, int l, int r, int p, int t) {
  if (l >= x && r <= y) {
    mn[p] = lazy[p] = t;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(p, l, r);
  lazy[p] = -1;
  if (mid >= x) update(x, y, l, mid, ls[p], t);
  if (mid < y) update(x, y, mid + 1, r, rs[p], t);
  mn[p] = min(mn[ls[p]], mn[rs[p]]);
}
int query(int x, int y, int l, int r, int p) {
  if (l >= x && r <= y) {
    return mn[p];
  }
  int mid = (l + r) >> 1;
  pushdown(p, l, r);
  lazy[p] = -1;
  int ans = 1e9;
  if (mid >= x) ans = min(ans, query(x, y, l, mid, ls[p]));
  if (mid < y) ans = min(ans, query(x, y, mid + 1, r, rs[p]));
  mn[p] = min(mn[ls[p]], mn[rs[p]]);
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &q);
  init();
  int t, l, r, x;
  mn[1] = RMQ(1, n);
  while (q--) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d", &x);
      update(l, r, 1, n * k, 1, x);
    } else {
      printf("%d\n", query(l, r, 1, n * k, 1));
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

You are given an array a consisting of positive integers and q queries to this array. There are two types of queries: 

  * 1 l r x — for each index i such that l ≤ i ≤ r set ai = x. 
  * 2 l r — find the minimum among such ai that l ≤ i ≤ r. 



We decided that this problem is too easy. So the array a is given in a compressed form: there is an array b consisting of n elements and a number k in the input, and before all queries a is equal to the concatenation of k arrays b (so the size of a is n·k).

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 104).

The second line contains n integers — elements of the array b (1 ≤ bi ≤ 109).

The third line contains one integer q (1 ≤ q ≤ 105).

Then q lines follow, each representing a query. Each query is given either as 1 l r x — set all elements in the segment from l till r (including borders) to x (1 ≤ l ≤ r ≤ n·k, 1 ≤ x ≤ 109) or as 2 l r — find the minimum among all elements in the segment from l till r (1 ≤ l ≤ r ≤ n·k).

Output

For each query of type 2 print the answer to this query — the minimum on the corresponding segment.

Examples

Input

3 1
1 2 3
3
2 1 3
1 1 2 4
2 1 3


Output

1
3


Input

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


Output

1
5
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int ls[N << 7], rs[N << 7];
int mn[N << 7], lazy[N << 7];
int Log[N], a[N], f[N][20];
int cnt = 1, q, n, k;
void init() {
  Log[0] = -1;
  for (int i = 1; i <= n; i++) f[i][0] = a[i], Log[i] = Log[i >> 1] + 1;
  for (int j = 1; j <= 19; j++)
    for (int i = 1; i + (1 << (j - 1)) <= n; i++)
      f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
int RMQ(int x, int y) {
  int t = Log[y - x + 1];
  return min(f[x][t], f[y - (1 << t) + 1][t]);
}
int qry(int l, int r) {
  if (r - l >= n)
    return RMQ(1, n);
  else {
    l = (l - 1) % n + 1, r = (r - 1) % n + 1;
    if (l > r)
      return min(RMQ(1, r), RMQ(l, n));
    else
      return RMQ(l, r);
  }
}
void pushdown(int p, int l, int r) {
  int mid = (l + r) >> 1;
  if (!ls[p]) ls[p] = ++cnt;
  if (!rs[p]) rs[p] = ++cnt;
  if (~lazy[p]) {
    lazy[ls[p]] = lazy[rs[p]] = lazy[p];
    mn[ls[p]] = lazy[ls[p]] ? lazy[ls[p]] : qry(l, mid);
    mn[rs[p]] = lazy[rs[p]] ? lazy[rs[p]] : qry(mid + 1, r);
  }
}
void update(int x, int y, int l, int r, int p, int t) {
  if (l >= x && r <= y) {
    mn[p] = lazy[p] = t;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(p, l, r);
  lazy[p] = -1;
  if (mid >= x) update(x, y, l, mid, ls[p], t);
  if (mid < y) update(x, y, mid + 1, r, rs[p], t);
  mn[p] = min(mn[ls[p]], mn[rs[p]]);
}
int query(int x, int y, int l, int r, int p) {
  if (l >= x && r <= y) {
    return mn[p];
  }
  int mid = (l + r) >> 1;
  pushdown(p, l, r);
  lazy[p] = -1;
  int ans = 1e9;
  if (mid >= x) ans = min(ans, query(x, y, l, mid, ls[p]));
  if (mid < y) ans = min(ans, query(x, y, mid + 1, r, rs[p]));
  mn[p] = min(mn[ls[p]], mn[rs[p]]);
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &q);
  init();
  int t, l, r, x;
  mn[1] = RMQ(1, n);
  while (q--) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d", &x);
      update(l, r, 1, n * k, 1, x);
    } else {
      printf("%d\n", query(l, r, 1, n * k, 1));
    }
  }
  return 0;
}

```