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;
inline long long fpow(long long a, long long x, long long p) {
  long long ans = 1;
  a %= p;
  while (x) {
    if (x & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    x >>= 1;
  }
  return ans;
}
struct Chtholly_Tree {
  long long l, r;
  mutable long long v;
  Chtholly_Tree(long long ls, long long rs = -1, long long vv = 0)
      : l(ls), r(rs), v(vv) {}
  bool operator<(const Chtholly_Tree& b) const { return l < b.l; }
};
set<Chtholly_Tree> s;
long long n, m, seed, vmax, op, l, r, x, y, a[100005];
inline long long rd() {
  long long ret = seed;
  seed = (seed * 7LL + 13LL) % 1000000007;
  return ret;
}
set<Chtholly_Tree>::iterator split(long long pos) {
  set<Chtholly_Tree>::iterator it = s.lower_bound(Chtholly_Tree(pos));
  if (it != s.end() && (*it).l == pos) return it;
  --it;
  if ((*it).r < pos) return s.end();
  long long ls = (*it).l, rs = (*it).r;
  long long vv = (*it).v;
  s.erase(it);
  s.insert(Chtholly_Tree(ls, pos - 1, vv));
  return s.insert(Chtholly_Tree(pos, rs, vv)).first;
}
void change(long long l, long long r, long long v) {
  set<Chtholly_Tree>::iterator itrs = split(r + 1LL), itls = split(l);
  s.erase(itls, itrs);
  s.insert(Chtholly_Tree(l, r, v));
}
void add(long long l, long long r, long long val) {
  set<Chtholly_Tree>::iterator itrs = split(r + 1LL), itls = split(l);
  for (; itls != itrs; ++itls) (*itls).v += val;
}
inline long long Rk_val(long long l, long long r, long long k) {
  vector<pair<long long, long long> > v;
  set<Chtholly_Tree>::iterator itrs = split(r + 1LL), itls = split(l);
  v.clear();
  for (; itls != itrs; ++itls)
    v.push_back(
        pair<long long, long long>((*itls).v, (*itls).r - (*itls).l + 1));
  sort(v.begin(), v.end());
  for (long long i = 0; i < v.size(); ++i) {
    k -= v[i].second;
    if (k <= 0) return v[i].first;
  }
  return -1LL;
}
inline long long POW(long long l, long long r, long long ex, long long p) {
  set<Chtholly_Tree>::iterator itrs = split(r + 1LL), itls = split(l);
  long long sum = 0LL;
  for (; itls != itrs; ++itls)
    sum = (sum + ((*itls).r - (*itls).l + 1LL) * fpow((*itls).v, x, p)) % p;
  return sum;
}
signed main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; ++i) {
    a[i] = rd() % vmax + 1;
    s.insert(Chtholly_Tree(i, i, a[i]));
  }
  s.insert(Chtholly_Tree(n + 1, n + 1, 0));
  for (long long i = 1; i <= m; ++i) {
    op = rd() % 4 + 1;
    l = rd() % n + 1;
    r = rd() % n + 1;
    if (l > r) swap(l, r);
    if (op == 3) {
      x = (rd() % (r - l + 1)) + 1;
      printf("%lld\n", Rk_val(l, r, x));
    } else
      x = (rd() % vmax) + 1;
    if (op == 4) {
      y = (rd() % vmax) + 1;
      printf("%lld\n", POW(l, r, x, y));
    }
    if (op == 1) add(l, r, x);
    if (op == 2) change(l, r, x);
  }
  return 0;
}