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;
const int maxn = 10e5 + 10;
int n, m;
long long seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
struct node {
  int l, r;
  mutable long long v;
  node(int _l, int _r = -1, long long _v = 0) : l(_l), r(_r), v(_v) {}
  bool operator<(const node &b) const { return this->l < b.l; }
};
set<node> ODT;
set<node>::iterator split(const int &pos) {
  set<node>::iterator p = ODT.lower_bound(node(pos));
  if (p != ODT.end() && p->l == pos) return p;
  --p;
  int ll = p->l, rr = p->r;
  long long v = p->v;
  ODT.erase(p);
  ODT.insert(node(ll, pos - 1, v));
  return ODT.insert(node(pos, rr, v)).first;
}
void assign(const int &l, const int &r, const long long &val) {
  set<node>::iterator p2 = split(r + 1), p1 = split(l);
  ODT.erase(p1, p2);
  ODT.insert(node(l, r, val));
}
void add(const int &l, const int &r, const long long &val) {
  set<node>::iterator p2 = split(r + 1), p1 = split(l);
  for (; p1 != p2; ++p1) p1->v += val;
}
long long findKth(const int &l, const int &r, int k) {
  set<node>::iterator p2 = split(r + 1), p1 = split(l);
  vector<pair<long long, int>> vec;
  for (; p1 != p2; ++p1) vec.push_back(make_pair(p1->v, p1->r - p1->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;
  }
  return -1LL;
}
long long fastpow(long long num, int k, int mod) {
  long long ans = 1, t = num;
  for (; k; k >>= 1) {
    if (k & 1) ans = (ans * t) % mod;
    t = (t % mod * t % mod) % mod;
  }
  return ans;
}
long long powSum(const int &l, const int &r, const int &mod, const int &pow) {
  set<node>::iterator p2 = split(r + 1), p1 = split(l);
  long long ans = 0;
  for (; p1 != p2; ++p1)
    ans =
        (ans + (long long)(p1->r - p1->l + 1) * fastpow(p1->v, pow, mod)) % mod;
  return ans;
}
int main() {
  scanf("%d%d%lld%lld", &n, &m, &seed, &vmax);
  for (int i = 1; i <= n; ++i) ODT.insert(node(i, i, (rnd() % vmax) + 1));
  ODT.insert(node(n + 1, n + 1, 0));
  int flag, l, r, x, y;
  for (int i = 1; i <= m; ++i) {
    flag = (rnd() % 4) + 1;
    l = (rnd() % n) + 1;
    r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (flag == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (flag == 4) y = (rnd() % vmax) + 1;
    switch (flag) {
      case 1:
        add(l, r, x);
        break;
      case 2:
        assign(l, r, x);
        break;
      case 3:
        printf("%lld\n", findKth(l, r, x));
        break;
      case 4:
        printf("%lld\n", powSum(l, r, y, x));
        break;
    }
  }
  return 0;
}