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;
long long n, m, seed, vmax, op, x, y;
long long a[1000001];
struct node {
  long long l, r;
  mutable long long v;
  node(long long ll, long long rr = -1, long long vv = 0)
      : l(ll), r(rr), v(vv) {}
  bool operator<(const node &u) const { return l < u.l; }
};
set<node> se;
long long poww(long long p, long long q, long long r) {
  long long res = 1, ans = p % r;
  while (q) {
    if (q & 1) res = res * ans % r;
    ans = ans * ans % r;
    q >>= 1;
  }
  return res;
}
set<node>::iterator sp(long long op) {
  set<node>::iterator it = se.lower_bound(node(op));
  if (it != se.end() && it->l == op) return it;
  it--;
  long long ll = it->l, rr = it->r, vv = it->v;
  se.erase(it);
  se.insert(node(ll, op - 1, vv));
  return se.insert(node(op, rr, vv)).first;
}
void ass(long long ll, long long rr, long long vv) {
  set<node>::iterator ir = sp(rr + 1), il = sp(ll);
  se.erase(il, ir);
  se.insert(node(ll, rr, vv));
}
void add(long long ll, long long rr, long long vv) {
  set<node>::iterator itr = sp(rr + 1), itl = sp(ll);
  for (; itl != itr; ++itl) {
    itl->v += vv;
  }
}
long long kth(long long ll, long long rr, long long k) {
  vector<pair<long long, long long> > ve;
  set<node>::iterator itr = sp(rr + 1), itl = sp(ll);
  for (; itl != itr; ++itl) {
    ve.push_back(pair<long long, long long>(itl->v, itl->r - itl->l + 1));
  }
  sort(ve.begin(), ve.end());
  for (vector<pair<long long, long long> >::iterator it = ve.begin();
       it != ve.end(); ++it) {
    k -= it->second;
    if (k <= 0) return it->first;
  }
  return 0;
}
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long mcf(long long l, long long r, long long x, long long y) {
  set<node>::iterator itr = sp(r + 1), itl = sp(l);
  long long res = 0;
  for (; itl != itr; ++itl) {
    res = (res + (itl->r - itl->l + 1) * poww(itl->v, x, y) % y) % y;
  }
  return res;
}
signed main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; ++i) {
    a[i] = rnd() % vmax + 1;
    se.insert(node(i, i, a[i]));
  }
  for (long long i = 1; i <= m; ++i) {
    long long op = (rnd() % 4) + 1, l = (rnd() % n) + 1, r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1) {
      add(l, r, x);
    }
    if (op == 2) {
      ass(l, r, x);
    }
    if (op == 3) {
      printf("%lld\n", kth(l, r, x));
    }
    if (op == 4) {
      printf("%lld\n", mcf(l, r, x, y));
    }
  }
  return 0;
}