Problem: — Willem...

— What's the matter?

— It seems that there's something wrong with Seniorious...

— I'll have a look...

<image>

Seniorious is made by linking special talismans in particular order.

After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

  * 1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
  * 2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
  * 3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It's guaranteed that 1 ≤ x ≤ r - l + 1.
  * 4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. <image>.

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:
    
    
      
    def rnd():  
      
        ret = seed  
        seed = (seed * 7 + 13) mod 1000000007  
        return ret  
      
    for i = 1 to n:  
      
        a[i] = (rnd() mod vmax) + 1  
      
    for i = 1 to m:  
      
        op = (rnd() mod 4) + 1  
        l = (rnd() mod n) + 1  
        r = (rnd() mod n) + 1  
      
        if (l > r):   
             swap(l, r)  
      
        if (op == 3):  
            x = (rnd() mod (r - l + 1)) + 1  
        else:  
            x = (rnd() mod vmax) + 1  
      
        if (op == 4):  
            y = (rnd() mod vmax) + 1  
      
    

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

Examples

Input

10 10 7 9


Output

2
1
0
3


Input

10 10 9 9


Output

1
1
3
3

Note

In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.

The operations are:

  * 2 6 7 9
  * 1 3 10 8
  * 4 4 6 2 4
  * 1 4 5 8
  * 2 1 7 1
  * 4 7 9 4 4
  * 1 2 7 9
  * 4 5 8 1 1
  * 2 5 7 5
  * 4 3 10 8 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n, m, vmax, tp;
pair<long long, int> st[N];
map<int, long long> M;
long long Pow(long long x, int b, int p, long long re = 1) {
  for (x %= p; b; b >>= 1, x = x * x % p)
    if (b & 1) re = re * x % p;
  return re;
}
namespace R {
const int mod = 1e9 + 7;
int seed, ret;
int rnd() {
  ret = seed;
  seed = (seed * 7LL + 13) % mod;
  return ret;
}
}  // namespace R
int main() {
  scanf("%d%d%d%d", &n, &m, &R::seed, &vmax);
  for (int i = 1; i <= n; ++i) M[i] = R::rnd() % vmax + 1;
  M[n + 1] = 0;
  for (int op, l, r, x, y; m; --m) {
    op = (R::rnd() & 3) + 1;
    l = R::rnd() % n + 1;
    r = R::rnd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = R::rnd() % (r - l + 1) + 1;
    else
      x = R::rnd() % vmax + 1;
    if (op == 4) y = R::rnd() % vmax + 1;
    auto p = --M.upper_bound(l);
    auto q = M.upper_bound(r);
    if (p->first < l) M[l] = p->second, ++p;
    if (r + 1 < q->first) --q, M[r + 1] = q->second, ++q;
    if (op == 1) {
      for (; p != q; ++p) p->second += x;
    }
    if (op == 2) {
      while (p != q) M.erase(p++);
      M[l] = x;
    }
    if (op == 3) {
      tp = 0;
      for (auto rin = p; p != q; ++p) {
        st[++tp] = {p->second, (++rin)->first - p->first};
      }
      sort(st + 1, st + 1 + tp);
      for (int i = 1; i <= tp; x -= st[i].second, ++i) {
        if (x <= st[i].second) {
          printf("%lld\n", st[i].first);
          break;
        }
      }
    }
    if (op == 4) {
      int ans = 0;
      for (auto rin = p; p != q; ++p) {
        ans = (ans + Pow(p->second, x, y) * ((++rin)->first - p->first)) % y;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}