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;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57)
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s * w;
}
const long long N = 1e5 + 5, P = 1e9 + 7;
struct ChthollyTree {
  map<long long, long long> s;
  inline long long find(long long p) { return s.lower_bound(p)->second; }
  inline void split(long long p) {
    long long pos = find(p);
    s[p] = pos;
  }
  inline pair<map<long long, long long>::iterator,
              map<long long, long long>::iterator>
  range(long long l, long long r) {
    split(l - 1), split(r);
    return make_pair(s.find(l - 1), s.find(r));
  }
} T;
long long n, m, seed, vmax, a[N], sum[N], len;
struct Node {
  long long val, cnt;
} A[N];
inline bool cmp(Node x, Node y) { return x.val < y.val; }
inline long long rnd() {
  long long res = seed;
  seed = (seed * 7ll + 13ll) % P;
  return res;
}
inline long long qpow(long long x, long long y, long long p) {
  long long ans = 1, res = x % p;
  while (y) {
    if (y & 1) ans = (ans * res) % p;
    y >>= 1;
    res = (res * res) % p;
  }
  return ans % p;
}
signed main() {
  n = read(), m = read(), seed = read(), vmax = read();
  for (register long long i = 1; i <= n; ++i) {
    a[i] = (rnd() % vmax) + 1;
    T.s[i] = a[i];
  }
  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;
    map<long long, long long>::iterator st, ed;
    if (op == 1) {
      pair<map<long long, long long>::iterator,
           map<long long, long long>::iterator>
          border = T.range(l, r);
      st = border.first, ed = border.second;
      while (st != ed) ed->second += x, ed--;
    } else if (op == 2) {
      pair<map<long long, long long>::iterator,
           map<long long, long long>::iterator>
          border = T.range(l, r);
      st = border.first, ed = border.second;
      while (st != ed) T.s.erase(ed--);
      T.s[r] = x;
    } else if (op == 3) {
      long long nxt;
      len = 0;
      for (register long long i = l; i <= r; i = nxt + 1) {
        map<long long, long long>::iterator now = T.s.lower_bound(i);
        nxt = min(r, now->first);
        A[++len] = (Node){now->second, nxt - i + 1};
      }
      sort(A + 1, A + len + 1, cmp);
      for (register long long i = 1; i <= len; ++i)
        sum[i] = sum[i - 1] + A[i].cnt;
      long long L = 1, R = len, res = -1;
      while (L <= R) {
        long long mid = (L + R) >> 1;
        if (sum[mid] >= x)
          res = mid, R = mid - 1;
        else
          L = mid + 1;
      }
      printf("%lld\n", A[res].val);
    } else {
      long long Ans = 0;
      long long nxt;
      for (register long long i = l; i <= r; i = nxt + 1) {
        map<long long, long long>::iterator now = T.s.lower_bound(i);
        nxt = min(r, now->first);
        Ans = (Ans + qpow(now->second, x, y) * (nxt - i + 1) % y) % y;
      }
      printf("%lld\n", Ans % y);
    }
  }
  return 0;
}