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 void R(int &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = (v + (v << 2) << 1) + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
inline void R(long long &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = (v + (v << 2) << 1) + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
long long seed, ret, vmax, n, m;
const long long mod = 1000000007;
inline long long rnd() {
  ret = seed;
  seed = ((long long)seed * 7 + 13) % mod;
  return ret;
}
long long a[100005];
map<long long, long long> q;
const long long inf = 0x3f3f3f3f;
pair<long long, long long> st[100005], st1[100005];
long long top, top1;
inline long long ksm(long long x, long long y, long long mod) {
  long long ret = 1;
  x %= mod;
  while (y) {
    if (y & 1) ret = (long long)ret * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ret;
}
long long ans;
signed main() {
  R(n), R(m), R(seed), R(vmax);
  for (long long i = 1; i <= n; ++i) {
    a[i] = rnd() % vmax + 1;
  }
  for (long long i = 1; i <= n; ++i) {
    q.insert(make_pair(i, a[i]));
  }
  long long kind, l, r, x, y;
  map<long long, long long>::iterator it1, it2;
  pair<long long, long long> temp;
  for (long long i = 1; i <= m; ++i) {
    kind = (rnd() % 4) + 1;
    l = (rnd() % n) + 1;
    r = (rnd() % n) + 1;
    if (l > r) {
      swap(l, r);
    }
    if (kind == 3) {
      x = (rnd() % (r - l + 1)) + 1;
    } else {
      x = (rnd() % vmax) + 1;
    }
    if (kind == 4) {
      y = (rnd() % vmax) + 1;
    }
    it1 = q.upper_bound(l);
    --it1;
    if (it1->first != l) {
      q[l] = it1->second;
      ++it1;
    }
    it2 = q.upper_bound(r + 1);
    --it2;
    if (it2->first != r + 1) {
      q[r + 1] = it2->second;
      ++it2;
    }
    switch (kind) {
      case 1:
        while (it1 != it2) {
          it1->second += x;
          ++it1;
        }
        break;
      case 2:
        while (it1 != it2) q.erase(it1++);
        q[l] = x;
        break;
      case 3: {
        top = 0;
        for (it1; it1->first <= r; ++it1) {
          st[++top] = *it1;
        }
        top1 = 0;
        for (long long i = 1; i < top; ++i) {
          st1[++top1] = make_pair(st[i].second, st[i + 1].first - st[i].first);
        }
        st1[++top1] = make_pair(st[top].second, r - st[top].first + 1);
        sort(st1 + 1, st1 + top1 + 1);
        long long now = 1;
        while (1) {
          if (x > st1[now].second) {
            x -= st1[now].second;
          } else {
            printf("%I64d\n", st1[now].first);
            break;
          }
          ++now;
        }
        break;
      }
      case 4:
        ans = 0;
        for (register long long ub; it1 != it2;)
          ub = it2--->first,
          ans = (ans + ksm(it2->second, x, y) * (ub - it2->first)) % y;
        printf("%I64d\n", ans);
        break;
    }
  }
  return 0;
}