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;
void readi(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void readll(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void readc(char &x) {
  char c;
  while ((c = getchar()) == ' ')
    ;
  x = c;
}
void writes(string s) { puts(s.c_str()); }
void writeln() { writes(""); }
void writei(int x) {
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
void writell(long long x) {
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
inline long long inc(int &x) { return ++x; }
inline long long inc(long long &x) { return ++x; }
inline long long inc(int &x, long long y) { return x += y; }
inline long long inc(long long &x, long long y) { return x += y; }
inline double inc(double &x, double y) { return x += y; }
inline long long dec(int &x) { return --x; }
inline long long dec(long long &x) { return --x; }
inline long long dec(int &x, long long y) { return x -= y; }
inline long long dec(long long &x, long long y) { return x -= y; }
inline double dec(double &x, double y) { return x -= y; }
inline long long mul(int &x) { return x = ((long long)x) * x; }
inline long long mul(long long &x) { return x = x * x; }
inline long long mul(int &x, long long y) { return x *= y; }
inline long long mul(long long &x, long long y) { return x *= y; }
inline double mul(double &x, double y) { return x *= y; }
inline long long divi(int &x) { return x = sqrt(x); }
inline long long divi(long long &x) { return x = sqrt(x); }
inline long long divi(int &x, long long y) { return x /= y; }
inline long long divi(long long &x, long long y) { return x /= y; }
inline double divi(double &x, double y) { return x /= y; }
inline long long mod(int &x, long long y) { return x %= y; }
inline long long mod(long long &x, long long y) { return x %= y; }
long long n, m, i, j, a[100005], seed, vmax, op, l, r, x, y;
set<pair<pair<long long, long long>, long long> > s;
vector<pair<pair<long long, long long>, long long> > v;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7ll + 13) % 1000000007;
  return ret;
}
void init() {
  long long i;
  if ((1) <= ((n)))
    for (((i)) = (1); ((i)) <= ((n)); ((i))++) a[i] = (rnd() % vmax) + 1;
}
void gt(long long &op, long long &l, long long &r, long long &x, long long &y) {
  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;
}
long long qp(long long x, long long y, long long p) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long z = qp(x, y / 2, p);
  z = z * z % p;
  if (y & 1) z = z * x % p;
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin >> n >> m >> seed >> vmax;
  init();
  if ((1) <= ((n)))
    for (((i)) = (1); ((i)) <= ((n)); ((i))++) {
      s.insert(make_pair(make_pair(i, i), a[i]));
    }
  while (m--) {
    gt(op, l, r, x, y);
    v.clear();
    set<pair<pair<long long, long long>, long long> >::iterator it =
        s.lower_bound(make_pair(make_pair(l, 0), 0));
    while (it != s.end() && it->first.second <= r) {
      v.push_back(*it);
      s.erase(it);
      it = s.lower_bound(make_pair(make_pair(l, 0), 0));
    }
    it = s.lower_bound(make_pair(make_pair(l, 0), 0));
    if (it != s.end() && it->first.first <= r) {
      v.push_back(make_pair(make_pair(it->first.first, r), it->second));
      pair<pair<long long, long long>, long long> t = *it;
      s.erase(it);
      s.insert(make_pair(make_pair(r + 1, t.first.second), t.second));
    }
    it = s.lower_bound(make_pair(make_pair(l, 0), 0));
    if (it != s.begin()) {
      it--;
      if (it->first.second >= l) {
        v.push_back(
            make_pair(make_pair(l, min(r, it->first.second)), it->second));
        pair<pair<long long, long long>, long long> t = *it;
        s.erase(it);
        s.insert(make_pair(make_pair(t.first.first, l - 1), t.second));
        if (r < t.first.second) {
          s.insert(make_pair(make_pair(r + 1, t.first.second), t.second));
        }
      }
    }
    if (op == 1) {
      for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
        s.insert(make_pair(make_pair(it->first.first, it->first.second),
                           it->second + x));
      }
    }
    if (op == 2) {
      s.insert(make_pair(make_pair(l, r), x));
    }
    if (op == 3) {
      vector<pair<long long, long long> > v2;
      for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
        s.insert(*it);
        v2.push_back(
            make_pair(it->second, it->first.second - it->first.first + 1));
      }
      stable_sort((v2).begin(), (v2).end());
      long long s = 0, ans, i = 0;
      while (i < v2.size() && s < x) {
        s += v2[i].second;
        ans = v2[i].first;
        i++;
      }
      cout << ans << endl;
    }
    if (op == 4) {
      long long ans = 0;
      for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
        s.insert(*it);
        inc(ans, (it->first.second - it->first.first + 1) *
                     qp(it->second % y, x, y) % y);
        mod(ans, y);
      }
      cout << ans << endl;
    }
  }
  return 0;
}