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 long long N = 1e5 + 55;
const long long moo = 1e9 + 7;
long long ret, seed, op, l, r, x, y;
long long rnd() {
  ret = seed;
  seed = (1ll * seed * 7 % moo + 13) % moo;
  return ret;
}
long long n, m, mod, a[N];
struct node {
  long long l, r;
  mutable long long v;
  friend bool operator<(node x, node y) { return x.l < y.l; }
};
set<node> s;
long long ksm(long long x, long long k, long long mo) {
  long long base = 1;
  while (k) {
    if (k & 1) base = 1ll * base * x % mo;
    x = 1ll * x * x % mo;
    k >>= 1;
  }
  return base;
}
set<node>::iterator split(long long pos) {
  set<node>::iterator it;
  it = s.lower_bound(node{pos, 0, 0});
  if (it->l == pos and it != s.end()) return it;
  it--;
  long long l = it->l;
  long long r = it->r;
  long long val = it->v;
  s.erase(it);
  s.insert((node){l, pos - 1, val});
  it = s.insert((node){pos, r, val}).first;
  return it;
}
void assign(long long l, long long r, long long val) {
  set<node>::iterator it1, it2;
  it1 = split(l);
  it2 = split(r + 1);
  s.erase(it1, it2);
  s.insert((node){l, r, val});
}
void add(long long l, long long r, long long val) {
  set<node>::iterator it1, it2;
  it1 = split(l);
  it2 = split(r + 1);
  for (; it1 != it2; it1++) {
    it1->v += val;
  }
}
long long top;
struct stt {
  long long len, val;
} st[N];
long long cmp(stt x, stt y) { return x.val < y.val; }
long long kth(long long l, long long r, long long x) {
  set<node>::iterator it1, it2;
  it1 = split(l);
  it2 = split(r + 1);
  top = 0;
  for (; it1 != it2; it1++) {
    st[++top] = {it1->r - it1->l + 1, it1->v};
  }
  sort(st + 1, st + top + 1, cmp);
  for (long long k = 1; k <= top; k++) {
    if (x <= st[k].len) {
      return st[k].val;
    }
    x -= st[k].len;
  }
}
long long sum(long long l, long long r, long long x, long long mo) {
  set<node>::iterator it1, it2;
  it1 = split(l);
  it2 = split(r + 1);
  long long ans = 0;
  for (; it1 != it2; it1++) {
    ans = (ans + 1ll * ksm((it1->v) % mo, x, mo) * (it1->r - it1->l + 1) % mo) %
          mo;
  }
  return ans;
}
signed main() {
  cin >> n >> m >> seed >> mod;
  for (long long i = 1; i <= n; i++) {
    a[i] = (rnd() % mod) + 1;
    node z = {i, i, a[i]};
    s.insert(z);
  }
  s.insert(node{n + 1, n + 1, -1});
  long long js = 0;
  for (long long 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) {
      x = (rnd() % (r - l + 1)) + 1;
      js++;
    } else {
      x = (rnd() % mod) + 1;
    }
    if (op == 4 and js == 100) {
      new char;
    }
    if (op == 4) {
      y = (rnd() % mod) + 1;
      js++;
    }
    if (js == 158) {
      new char;
    }
    if (op == 1) add(l, r, x);
    if (op == 2) assign(l, r, x);
    if (op == 3) printf("%lld\n", kth(l, r, x));
    if (op == 4) printf("%lld\n", sum(l, r, x, y));
  }
  return 0;
}