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;
template <typename T>
bool chkmx(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmn(T &x, T y) {
  return x > y ? x = y, true : false;
}
const int maxn = 1e5;
int n, m, seed, vmx, opt, l, r, x, y;
struct Node {
  int l, r;
  mutable long long x;
  inline friend bool operator<(Node a, Node b) { return a.l < b.l; }
};
set<Node> s;
inline int rnd() {
  int ret = seed;
  seed = (1ll * seed * 7 + 13) % (int)(1e9 + 7);
  return ret;
}
inline int fpw(long long a, long long b, long long mod, long long ans = 1) {
  a %= mod;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
set<Node>::iterator split(int p) {
  set<Node>::iterator it = s.lower_bound((Node){p, 0, 0});
  if (it != s.end() && it->l == p) return it;
  it--;
  int l = it->l, r = it->r;
  long long x = it->x;
  s.erase(it), s.insert((Node){l, p - 1, x});
  return s.insert((Node){p, r, x}).first;
}
void add(int l, int r, int k) {
  set<Node>::iterator rit = split(r + 1), lit = split(l);
  for (set<Node>::iterator it = lit; it != rit; it++) it->x += k;
}
void cover(int l, int r, int k) {
  set<Node>::iterator rit = split(r + 1), lit = split(l);
  s.erase(lit, rit), s.insert((Node){l, r, k});
}
long long kth(int l, int r, int k) {
  vector<pair<long long, int> > v;
  set<Node>::iterator rit = split(r + 1), lit = split(l);
  for (set<Node>::iterator it = lit; it != rit; it++)
    v.push_back(make_pair(it->x, it->r - it->l + 1));
  sort(v.begin(), v.end());
  for (int i = (0); i <= int((int)(v.size()) - 1); i++) {
    k -= v[i].second;
    if (k <= 0) return v[i].first;
  }
}
int ksum(int l, int r, int k, int mod) {
  set<Node>::iterator rit = split(r + 1), lit = split(l);
  int ans = 0;
  for (set<Node>::iterator it = lit; it != rit; it++)
    ans = (ans + 1ll * (it->r - it->l + 1) * fpw(it->x, k, mod) % mod) % mod;
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &seed, &vmx);
  for (int i = (1); i <= int(n); i++) s.insert((Node){i, i, rnd() % vmx + 1});
  for (int i = (1); i <= int(m); i++) {
    opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1;
    if (l > r) swap(l, r);
    if (opt == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmx + 1;
    if (opt == 4) y = rnd() % vmx + 1;
    if (opt == 1)
      add(l, r, x);
    else if (opt == 2)
      cover(l, r, x);
    else if (opt == 3)
      printf("%lld\n", kth(l, r, x));
    else if (opt == 4)
      printf("%d\n", ksum(l, r, x, y));
  }
  return 0;
}