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 double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {
  if (fabs(x) <= eps) return 0;
  return (x > 0) ? 1 : -1;
};
void umax(long long &a, long long b) { a = max(a, b); }
void umin(long long &a, long long b) { a = min(a, b); }
void file() {
  freopen("out.txt", "r", stdin);
  freopen("1.txt", "w", stdout);
}
namespace solver {
long long Pow(long long a, long long b, long long mod) {
  a %= mod;
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res %= mod;
    b >>= 1;
    a *= a;
    a %= mod;
  }
  return res;
}
namespace Chtholly_TREE {
struct node {
  long long l, r;
  mutable long long v;
  node(long long L, long long R = -1, long long V = 0) : l(L), r(R), v(V) {}
  bool operator<(const node &b) const { return l < b.l; }
};
set<node> st;
set<node>::iterator split(long long pos) {
  auto it = st.lower_bound(node(pos));
  if (it != st.end() && it->l == pos) return it;
  --it;
  if (pos > it->r) return st.end();
  long long L = it->l, R = it->r;
  long long v = it->v;
  st.erase(it);
  st.insert(node(L, pos - 1, v));
  return st.insert(node(pos, R, v)).first;
}
void init(int n) {
  st.clear();
  st.insert(node(0, 2e9, 0));
}
void add(long long l, long long r, long long val = 1) {
  split(l);
  auto itr = split(r + 1), itl = split(l);
  for (; itl != itr; ++itl) itl->v += val;
}
void cover(long long l, long long r, long long val = 0) {
  split(l);
  auto itr = split(r + 1), itl = split(l);
  st.erase(itl, itr);
  st.insert(node(l, r, val));
}
vector<pair<long long, long long>> V;
long long askth(long long l, long long r, long long k, bool reversed = 0) {
  if (reversed) k = r - l + 2 - k;
  split(l);
  auto itr = split(r + 1), itl = split(l);
  V.clear();
  for (; itl != itr; ++itl) V.push_back({itl->v, itl->r - itl->l + 1});
  sort(V.begin(), V.end());
  for (auto it : V) {
    k -= it.second;
    if (k <= 0) return it.first;
  }
  return -1;
}
long long sum(long long l, long long r, long long ex, long long mod) {
  split(l);
  auto itr = split(r + 1), itl = split(l);
  long long res = 0;
  for (; itl != itr; ++itl)
    res += (itl->r - itl->l + 1) * Pow(itl->v, ex, mod), res %= mod;
  return res;
}
};  // namespace Chtholly_TREE
using namespace Chtholly_TREE;
long long n, m, seed, vmax;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
long long a[110000];
void solve() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &vmax);
  init(n);
  for (long long i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1, add(i, i, a[i]);
  for (long long i = 1; i <= m; i++) {
    long long op = (rnd() % 4) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    if (op == 3) {
      long long x = (rnd() % (r - l + 1)) + 1;
      printf("%lld\n", askth(l, r, x));
      continue;
    }
    long long x = (rnd() % vmax) + 1;
    if (op == 1)
      add(l, r, x);
    else if (op == 2)
      cover(l, r, x);
    if (op == 4) {
      long long y = (rnd() % vmax) + 1;
      printf("%lld\n", sum(l, r, x, y));
    }
  }
}
}  // namespace solver
int main() {
  solver::solve();
  return 0;
}