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;
struct node {
  long long l, r;
  mutable long long v;
  node() {}
  node(long long _l, long long _r, long long _v) : l(_l), r(_r), v(_v) {}
  const bool operator<(const node& x) const { return l < x.l; }
};
set<node> s;
long long n, m, seed, vmax, op, l, r, x, y;
long long a[100005];
long long rnd(long long mod) {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret % mod + 1;
}
long long fpow(long long a, long long b, long long mod) {
  long long res = 1, ans = a % mod;
  while (b) {
    if (b & 1) res = res * ans % mod;
    ans = ans * ans % mod;
    b >>= 1;
  }
  return res;
}
set<node>::iterator split(long long pos) {
  set<node>::iterator it = s.lower_bound(node(pos, 0, 0));
  if (it != s.end() && it->l == pos) return it;
  it--;
  long long l = it->l, r = it->r, v = it->v;
  s.erase(it);
  s.insert(node(l, pos - 1, v));
  return s.insert(node(pos, r, v)).first;
}
void assign(long long l, long long r, long long v) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  s.erase(itl, itr);
  s.insert(node(l, r, v));
}
void add(long long l, long long r, long long x) {
  set<node>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; itl++) itl->v += x;
}
long long kth(long long l, long long r, long long k) {
  vector<pair<long long, long long> > vec;
  set<node>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; itl++)
    vec.push_back(make_pair(itl->v, itl->r - itl->l + 1));
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    k -= vec[i].second;
    if (k <= 0) return vec[i].first;
  }
}
long long sum(long long l, long long r, long long x, long long y) {
  long long res = 0;
  set<node>::iterator itr = split(r + 1), itl = split(l);
  for (; itl != itr; itl++)
    res = (res + (itl->r - itl->l + 1) % y * fpow(itl->v, x, y) % y) % y;
  return res;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; i++) {
    a[i] = rnd(vmax);
    s.insert(node(i, i, a[i]));
  }
  for (int i = 1; i <= m; i++) {
    op = rnd(4), l = rnd(n), r = rnd(n);
    if (l > r) swap(l, r);
    if (op == 3)
      x = rnd(r - l + 1);
    else
      x = rnd(vmax);
    if (op == 4) y = rnd(vmax);
    switch (op) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        assign(l, r, x);
        break;
      case 3:
        printf("%lld\n", kth(l, r, x));
        break;
      case 4:
        printf("%lld\n", sum(l, r, x, y));
        break;
    }
  }
  return 0;
}