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 = 1e5 + 7;
int a[MAXN];
int n, m, seed, vmax;
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& o) const { return l < o.l; }
};
set<NODE> s;
vector<pair<long long, int> > vp;
set<NODE>::iterator split(int pos) {
  auto it = s.lower_bound(NODE(pos));
  if (it != s.end() && it->l == pos) return it;
  --it;
  if (pos > it->r) return s.end();
  int L = it->l, R = it->r;
  long long V = it->v;
  s.erase(it);
  s.insert(NODE(L, pos - 1, V));
  return s.insert(NODE(pos, R, V)).first;
}
void add(int l, int r, long long val = 1) {
  split(l);
  auto itr = split(r + 1), itl = split(l);
  for (; itl != itr; ++itl) itl->v += val;
}
void assign(int l, int r, long long val = 0) {
  split(l);
  auto itr = split(r + 1), itl = split(l);
  s.erase(itl, itr);
  s.insert(NODE(l, r, val));
}
long long rnk(int l, int r, int k, bool reversed = 0) {
  if (reversed) k = r - l + 2 - k;
  split(l);
  auto itr = split(r + 1), itl = split(l);
  vp.clear();
  for (; itl != itr; ++itl) vp.push_back({itl->v, itl->r - itl->l + 1});
  sort(vp.begin(), vp.end());
  for (auto i : vp) {
    k -= i.second;
    if (k <= 0) return i.first;
  }
  return -1;
}
long long qpow(long long n, long long k, long long p) {
  long long res = 1;
  for (; k; n = n * n % p, k >>= 1) {
    if (k & 1) res = res * n % p;
  }
  return res;
}
long long sum(int l, int r, int ex, int mod) {
  split(l);
  auto itr = split(r + 1), itl = split(l);
  long long res = 0;
  for (; itl != itr; ++itl)
    res =
        (res + (long long)(itl->r - itl->l + 1) * qpow(itl->v % mod, ex, mod)) %
        mod;
  return res;
}
int rnd() {
  int ret = seed;
  seed = ((long long)seed * 7 + 13) % 1000000007;
  return ret;
}
void pre() {
  int l = 1, r = 1;
  for (int i = 2; i <= n + 1; i++) {
    if (a[i] != a[i - 1]) {
      r = i - 1;
      s.insert(NODE(l, r, a[i - 1]));
      l = i;
    }
  }
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; i++) {
    a[i] = (rnd() % vmax) + 1;
  }
  pre();
  int op, l, r, k, p;
  for (int i = 1; i <= m; i++) {
    op = (rnd() % 4) + 1;
    l = (rnd() % n) + 1;
    r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (op == 3)
      k = (rnd() % (r - l + 1)) + 1;
    else
      k = (rnd() % vmax) + 1;
    if (op == 4) p = (rnd() % vmax) + 1;
    if (op == 1) {
      add(l, r, k);
    }
    if (op == 2) {
      assign(l, r, k);
    }
    if (op == 3) {
      cout << rnk(l, r, k) << endl;
    }
    if (op == 4) {
      cout << sum(l, r, k, p) << endl;
    }
  }
  return 0;
}