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 val;
  Node(long long l, long long r, long long val) : l(l), r(r), val(val) {}
  inline bool operator<(const Node b) const { return l < b.l; }
};
set<Node> Ctholly;
long long n, m, seed, vmax;
long long a[100010];
long long fpow(long long a, long long b, long long p) {
  a %= p;
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return res;
}
set<Node>::iterator split(long long x) {
  set<Node>::iterator it = Ctholly.lower_bound(Node(x, 0, 0));
  if (it->l == x && it != Ctholly.end()) return it;
  --it;
  long long l = it->l, r = it->r, v = it->val;
  Ctholly.erase(it);
  Ctholly.insert(Node(l, x - 1, v));
  return Ctholly.insert(Node(x, r, v)).first;
}
void assign(long long l, long long r, long long x) {
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  Ctholly.erase(itl, itr);
  Ctholly.insert(Node(l, r, x));
}
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
void add(long long l, long long r, long long x) {
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  for (set<Node>::iterator it = itl; it != itr; it++) it->val += x;
}
void ssort(long long l, long long r, long long x) {
  static set<Node>::iterator arr[100010];
  static function<long long(set<Node>::iterator)> fun =
      [](set<Node>::iterator x) { return x->r - x->l + 1; };
  long long cnt = 0;
  set<Node>::iterator itr = split(r + 1), itl = split(l);
  for (set<Node>::iterator it = itl; it != itr; it++) arr[++cnt] = it;
  sort(arr + 1, arr + cnt + 1,
       [](set<Node>::iterator a, set<Node>::iterator b) {
         return a->val < b->val;
       });
  for (long long i = 1; i <= cnt; i++) {
    if (fun(arr[i]) >= x) {
      printf("%lld\n", arr[i]->val);
      return;
    } else
      x -= fun(arr[i]);
  }
}
void 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 (set<Node>::iterator it = itl; it != itr; it++)
    res = (res + fpow(it->val, x, y) * (it->r - it->l + 1) % y) % y;
  printf("%lld\n", res);
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; i++) {
    long long tmp = rnd() % vmax + 1;
    Ctholly.insert(Node(i, i, tmp));
  }
  Ctholly.insert(Node(n + 1, n + 1, 0));
  while (m--) {
    long long opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x,
              y = -1;
    if (l > r) swap(l, r);
    if (opt == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % vmax + 1;
    if (opt == 4) y = rnd() % vmax + 1;
    if (opt == 1)
      add(l, r, x);
    else if (opt == 2)
      assign(l, r, x);
    else if (opt == 3)
      ssort(l, r, x);
    else
      sum(l, r, x, y);
  }
}