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 N = 100005;
int n, m;
long long seed, vmax;
int opt, l, r, x, y;
struct node {
  int l, r;
  mutable long long v;
  bool operator<(const node &a) const { return l < a.l; }
  node(int a = 0, int b = 0, long long c = 0) {
    l = a;
    r = b;
    v = c;
  }
};
struct num {
  int t;
  long long v;
} a[N];
set<node> s;
long long rd() {
  long long t = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return t;
}
set<node>::iterator split(int pos) {
  set<node>::iterator it = s.lower_bound(node(pos, pos, -1));
  if (it != s.end() && it->l == pos) return it;
  --it;
  int lr = it->l, rr = it->r;
  long long v = it->v;
  s.erase(it);
  s.insert(node(lr, pos - 1, v));
  return s.insert(node(pos, rr, v)).first;
}
void assign(int l, int r, long long v) {
  set<node>::iterator ir = split(r + 1);
  set<node>::iterator il = split(l);
  s.erase(il, ir);
  s.insert(node(l, r, v));
}
void add(int l, int r, long long v) {
  set<node>::iterator ir = split(r + 1);
  set<node>::iterator il = split(l);
  for (set<node>::iterator x = il; x != ir; ++x) x->v += v;
}
bool cmp(num a, num b) { return a.v < b.v; }
long long kth(int l, int r, int k) {
  set<node>::iterator itr = split(r + 1);
  set<node>::iterator itl = split(l);
  int tp = 0;
  for (set<node>::iterator x = itl; x != itr; ++x) {
    a[++tp].t = (x->r - x->l + 1);
    a[tp].v = x->v;
  }
  sort(a + 1, a + 1 + tp, cmp);
  for (register int i = 1; i <= tp; ++i) {
    k -= a[i].t;
    if (k <= 0) return a[i].v;
  }
  return -1LL;
}
int qui(long long x, int b, int p) {
  long long s = 1;
  x %= p;
  while (b) {
    if (b & 1) s = 1LL * s * x % p;
    x = 1LL * x * x % p;
    b >>= 1;
  }
  return s;
}
long long xpow(int l, int r, int x, int y) {
  long long ans = 0;
  set<node>::iterator itr = split(r + 1);
  set<node>::iterator itl = split(l);
  for (set<node>::iterator it = itl; it != itr; ++it)
    ans = (ans + 1LL * (it->r - it->l + 1) * qui(it->v, x, y) % y) % y;
  return ans;
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (register int i = 1; i <= n; ++i) s.insert(node(i, i, rd() % vmax + 1));
  s.insert(node(n + 1, n + 1, 0));
  for (register int i = 1; i <= m; ++i) {
    opt = (rd() % 4) + 1;
    l = (rd() % n) + 1;
    r = (rd() % n) + 1;
    if (l > r) swap(l, r);
    if (opt == 3)
      x = (rd() % (r - l + 1)) + 1;
    else
      x = (rd() % vmax) + 1;
    if (opt == 4) y = (rd() % vmax) + 1;
    switch (opt) {
      case 1: {
        add(l, r, x);
        break;
      }
      case 2: {
        assign(l, r, x);
        break;
      }
      case 3: {
        printf("%lld\n", kth(l, r, x));
        break;
      }
      case 4: {
        printf("%lld\n", xpow(l, r, x, y));
        break;
      }
    }
  }
  return 0;
}