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;
long long n, m, seed, vmax, l, r, x, y;
long long mod = 1e9 + 7;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7LL + 13) % mod;
  return ret;
}
struct str {
  long long l, r;
  long long v;
  bool operator<(const str &o) const { return l < o.l || l == o.l && r < o.r; }
};
bool cmp(const str &a, const str &b) { return a.v < b.v; }
set<str> s;
long long a[100005];
long long powx(long long a) {
  long long ret = 1;
  a %= y;
  for (long long t = x; t; a = (a * a) % y, t >>= 1) {
    if (t & 1) {
      ret *= a;
      ret %= y;
    }
  }
  return ret;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &m, &seed, &vmax);
  for (long long i = 1; i <= n; i++) a[i] = (rnd() % vmax) + 1;
  for (long long i = 1; i <= n; i++) {
    long long cur = i;
    while (a[cur] == a[i + 1]) i++;
    s.insert({cur, i, a[i]});
  }
  s.insert({mod, mod, 0});
  for (long long i = 1; i <= m; i++) {
    y = 0;
    long long 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;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    auto begin = s.upper_bound({l, mod, 0});
    begin--;
    if (begin->l < l) {
      s.insert({begin->l, l - 1, begin->v});
      s.insert({l, begin->r, begin->v});
      s.erase(begin);
    }
    auto end = s.upper_bound({r, mod, 0});
    end--;
    if (r < end->r) {
      s.insert({end->l, r, end->v});
      s.insert({r + 1, end->r, end->v});
      s.erase(end);
    }
    begin = s.upper_bound({l, mod, 0});
    begin--;
    end = s.upper_bound({r, mod, 0});
    end--;
    if (op == 1) {
      for (auto it = begin; it->r <= r; it++) {
        str tmp = *it;
        s.erase(it);
        it = s.insert({tmp.l, tmp.r, tmp.v + x}).first;
      }
    } else if (op == 2) {
      s.erase(begin, ++end);
      s.insert({l, r, x});
    } else if (op == 3) {
      vector<str> v;
      for (auto it = begin; it->r <= end->r; it++) v.push_back(*it);
      sort(v.begin(), v.end(), cmp);
      for (str cur : v) {
        if (cur.r - cur.l + 1 < x)
          x -= cur.r - cur.l + 1;
        else {
          printf("%lld\n", cur.v);
          break;
        }
      }
    } else {
      long long ret = 0;
      for (auto it = begin; it->r <= end->r; it++) {
        ret += powx(it->v) * (it->r - it->l + 1);
        ret %= y;
      }
      printf("%lld\n", ret);
    }
  }
}