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 long long maxn = 100010, mod = 1e9 + 7;
struct node {
  long long l, r;
  mutable long long v;
  node(long long stdl, long long stdr = -1, long long stdv = 0)
      : l(stdl), r(stdr), v(stdv) {}
  bool operator<(const node &x) const { return l < x.l; }
};
long long qpow(long long x, long long y, long long p) {
  long long res = 1;
  long long ans = x % p;
  while (y) {
    if (y & 1) res = res * ans % p;
    ans = ans * ans % p;
    y >>= 1;
  }
  return res;
}
set<node> s;
set<node>::iterator split(long long x) {
  set<node>::iterator it = s.lower_bound(node(x));
  if (it != s.end() && it->l == x) return it;
  it--;
  long long stdl = it->l, stdr = it->r, stdv = it->v;
  s.erase(it);
  s.insert(node(stdl, x - 1, stdv));
  return s.insert(node(x, stdr, stdv)).first;
}
void assign(long long l, long long r, long long k) {
  set<node>::iterator stdr = split(r + 1), stdl = split(l);
  s.erase(stdl, stdr);
  s.insert(node(l, r, k));
}
void add(long long l, long long r, long long k) {
  set<node>::iterator stdr = split(r + 1), stdl = split(l);
  for (; stdl != stdr; ++stdl) stdl->v += k;
}
long long query_rank(long long l, long long r, long long k) {
  vector<pair<long long, long long> > vp;
  set<node>::iterator stdr = split(r + 1), stdl = split(l);
  vp.clear();
  for (; stdl != stdr; ++stdl)
    vp.push_back(pair<long long, long long>(stdl->v, stdl->r - stdl->l + 1));
  sort(vp.begin(), vp.end());
  for (vector<pair<long long, long long> >::iterator it = vp.begin();
       it != vp.end(); it++) {
    k -= it->second;
    if (k <= 0) return it->first;
  }
}
long long query_sum(long long l, long long r, long long k, long long p) {
  set<node>::iterator stdr = split(r + 1), stdl = split(l);
  long long res = 0;
  for (; stdl != stdr; ++stdl)
    res += (stdl->r - stdl->l + 1) * qpow(stdl->v, k, p), res %= p;
  return res;
}
long long n, m, seed, vmax, a[maxn];
long long rd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % mod;
  return ret;
}
signed main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; i++)
    a[i] = rd() % vmax + 1, s.insert(node(i, i, a[i]));
  s.insert(node(n + 1, n + 1, 0));
  while (m--) {
    long long opt = (rd() % 4) + 1, l = (rd() % n) + 1, r = (rd() % n) + 1;
    if (l > r) swap(l, r);
    long long x, y;
    if (opt == 3)
      x = rd() % (r - l + 1) + 1;
    else
      x = rd() % vmax + 1;
    if (opt == 4) y = rd() % vmax + 1;
    if (opt == 1) add(l, r, x);
    if (opt == 2) assign(l, r, x);
    if (opt == 3) printf("%lld\n", query_rank(l, r, x));
    if (opt == 4) printf("%lld\n", query_sum(l, r, x, y));
  }
  return 0;
}