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 {
  int l, r;
  mutable long long val;
  node(int L = -1, int R = -1, long long Val = 0) { l = L, r = R, val = Val; }
  bool operator<(const node &d) const { return l < d.l; }
};
set<node> s;
int n, m, inp;
int seed, vmax;
vector<pair<long long, int> > v;
int rnd() {
  int ret = seed;
  seed = ((long long)seed * 7LL + 13LL) % 1000000007;
  return ret;
}
set<node>::iterator split(int pos) {
  set<node>::iterator it = s.lower_bound(node(pos));
  if (it != s.end() && it->l == pos) return it;
  it--;
  int L = it->l, R = it->r;
  long long V = it->val;
  s.erase(it);
  s.insert(node(L, pos - 1, V));
  return s.insert(node(pos, R, V)).first;
}
void merge(int l, int r, int v) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  s.erase(itl, itr);
  s.insert(node(l, r, (long long)v));
}
void add(int l, int r, int v) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  for (; itl != itr; itl++) itl->val += v;
}
long long find_kth(int l, int r, int k) {
  v.clear();
  set<node>::iterator itl = split(l), itr = split(r + 1);
  for (; itl != itr; itl++)
    v.push_back(make_pair(itl->val, (itl->r) - (itl->l) + 1));
  sort(v.begin(), v.end());
  for (int i = 0; i <= (int)v.size() - 1; i++) {
    k -= v[i].second;
    if (k <= 0) return v[i].first;
  }
  return -1LL;
}
int power(int a, int x, int p) {
  int sum = 1;
  while (x) {
    if (x & 1) sum = (long long)sum * a % p;
    a = (long long)a * a % p;
    x >>= 1;
  }
  return sum;
}
int get_sum(int l, int r, int x, int p) {
  set<node>::iterator itl = split(l), itr = split(r + 1);
  int sum = 0;
  for (; itl != itr; itl++)
    sum = ((long long)sum +
           (long long)((itl->r) - (itl->l) + 1) * power(itl->val % p, x, p)) %
          p;
  return sum;
}
int main() {
  int opt, l, r, x, y;
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) {
    inp = rnd() % vmax + 1;
    s.insert(node(i, i, (long long)inp));
  }
  for (int i = 1; i <= 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() % vmax + 1;
    if (opt == 4) y = rnd() % vmax + 1;
    switch (opt) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        merge(l, r, x);
        break;
      case 3:
        printf("%lld\n", find_kth(l, r, x));
        break;
      case 4:
        printf("%d\n", get_sum(l, r, x, y));
        break;
    }
  }
  return 0;
}