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>
#pragma GCC optimize(2)
using namespace std;
const long long N = 100010, mod = 1000000007;
long long n, m, seed, vmax, a[N];
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  long long y = 10, len = 1;
  while (y <= x) {
    y = (y << 3) + (y << 1);
    len++;
  }
  while (len--) {
    y /= 10;
    putchar(x / y + 48);
    x %= y;
  }
}
long long ksm(long long x, long long y, long long mod) {
  if (y == 0) return 1;
  long long t = ksm(x, y / 2, mod) % mod;
  t = t * t % mod;
  if (y & 1) t = t * x % mod;
  return t;
}
inline long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % mod;
  return ret;
}
namespace ODT {
struct node {
  long long l, r;
  mutable long long val;
  node(long long L, long long R = -1, long long Val = 0)
      : l(L), r(R), val(Val) {}
  inline bool operator<(const node &x) const { return l < x.l; }
};
set<node> s;
inline set<node>::iterator split(long long x) {
  set<node>::iterator i = s.lower_bound(node(x));
  if (i != s.end() && i->l == x) return i;
  i--;
  long long l = i->l, r = i->r, val = i->val;
  s.erase(i);
  s.insert(node(l, x - 1, val));
  return s.insert(node(x, r, val)).first;
}
inline void update(long long l, long long r, long long val) {
  for (set<node>::iterator L = split(l), R = split(r + 1); L != R; L++)
    L->val += val;
}
inline void giveval(long long l, long long r, long long val) {
  set<node>::iterator L = split(l), R = split(r + 1);
  s.erase(L, R);
  s.insert(node(l, r, val));
}
inline long long getvalbyrank(long long l, long long r, long long rank) {
  vector<pair<long long, long long> > G;
  G.clear();
  for (set<node>::iterator L = split(l), R = split(r + 1); L != R; L++)
    G.push_back(make_pair(L->val, L->r - L->l + 1));
  sort(G.begin(), G.end());
  for (long long i = 0; i < G.size(); i++) {
    rank -= G[i].second;
    if (rank <= 0) return G[i].first;
  }
  return -1;
}
inline long long getsum(long long l, long long r, long long ex, long long mod) {
  long long ans = 0;
  for (set<node>::iterator L = split(l), R = split(r + 1); L != R; L++)
    ans = (ans + (L->r - L->l + 1) * ksm(L->val % mod, ex, mod)) % mod;
  return ans;
}
}  // namespace ODT
int main() {
  cin >> n >> m >> seed >> vmax;
  for (long long i = 1; i <= n; i++) {
    a[i] = rnd() % vmax + 1;
    ODT::s.insert(ODT::node{i, i, a[i]});
  }
  ODT::s.insert(ODT::node{n + 1, n + 1, 0});
  while (m--) {
    long long op = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    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;
    switch (op) {
      case 1: {
        ODT::update(l, r, x);
        break;
      }
      case 2: {
        ODT::giveval(l, r, x);
        break;
      }
      case 3: {
        write(ODT::getvalbyrank(l, r, x));
        putchar('\n');
        break;
      }
      case 4: {
        write(ODT::getsum(l, r, x, y));
        putchar('\n');
        break;
      }
    }
  }
  return 0;
}