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 maxn = 100005;
const long long mod = 1000000007;
long long i, j, k, m, n;
long long seed, maxx;
long long a[maxn];
inline void swp(long long &a, long long &b) { a += b, b = a - b, a -= b; }
long long ksm(long long a, long long b, long long p) {
  a %= p;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p, b >>= 1;
  }
  return res;
}
struct Chtholly_Tree {
  struct section {
    long long l, r;
    mutable long long v;
    section(long long L, long long R, long long V) { l = L, r = R, v = V; }
    bool operator<(const section x) const { return l < x.l; }
  };
  set<section> s;
  void build(long long n) {
    for (long long i = 1; i <= n; i++) s.insert(section(i, i, a[i]));
  }
  set<section>::iterator split(long long k) {
    set<section>::iterator it = s.lower_bound(section(k, -1, 0));
    if (it->l == k) return it;
    long long l, r, v;
    it--;
    l = it->l, r = it->r, v = it->v;
    s.erase(it);
    s.insert(section(l, k - 1, v));
    return s.insert(section(k, r, v)).first;
  }
  void assign(long long L, long long R, long long V) {
    set<section>::iterator r = split(R + 1), l = split(L);
    s.erase(l, r);
    s.insert(section(L, R, V));
  }
  void add(long long L, long long R, long long V) {
    set<section>::iterator r = split(R + 1), l = split(L);
    while (l != r) {
      l->v += V;
      l++;
    }
  }
  long long getkth(long long L, long long R, long long k) {
    vector<pair<long long, long long> > v;
    set<section>::iterator r = split(R + 1), l = split(L);
    vector<pair<long long, long long> >::iterator it;
    v.clear();
    while (l != r) {
      v.push_back(make_pair(l->v, l->r - l->l + 1));
      l++;
    }
    sort(v.begin(), v.end());
    for (it = v.begin(); it != v.end(); it++) {
      k -= it->second;
      if (k <= 0) return it->first;
    }
  }
  long long getsum(long long L, long long R, long long x, long long y) {
    long long res = 0;
    set<section>::iterator r = split(R + 1), l = split(L);
    while (l != r) {
      res = (res + ksm(l->v, x, y) * (l->r - l->l + 1)) % y;
      l++;
    }
    return res;
  }
} ODT;
inline long long rnd() {
  long long res = seed;
  seed = (seed * 7 + 13) % mod;
  return res;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &m, &seed, &maxx);
  for (i = 1; i <= n; i++) a[i] = rnd() % maxx + 1;
  ODT.build(n);
  for (i = 1; i <= m; i++) {
    long long opt = rnd() % 4 + 1, l = rnd() % n + 1, r = rnd() % n + 1, x, y;
    if (l > r) swp(l, r);
    if (opt == 3)
      x = rnd() % (r - l + 1) + 1;
    else
      x = rnd() % maxx + 1;
    if (opt == 4) y = rnd() % maxx + 1;
    if (opt == 1) ODT.add(l, r, x);
    if (opt == 2) ODT.assign(l, r, x);
    if (opt == 3) printf("%lld\n", ODT.getkth(l, r, x));
    if (opt == 4) printf("%lld\n", ODT.getsum(l, r, x, y));
  }
  return 0;
}