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 int inf = 0x3f3f3f3f;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
template <typename T>
inline void umax(T& a, T b) {
  a = max(a, b);
}
template <typename T>
inline void umin(T& a, T b) {
  a = min(a, b);
}
void debug() { cout << endl; }
template <typename T, typename... R>
void debug(T f, R... r) {
  cout << "[" << f << "]";
  debug(r...);
}
const int MAXN = 100005;
long long n, m, seed, vmax, a[MAXN];
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % 1000000007;
  return ret;
}
struct Line {
  pair<int, int> itv;
  long long val;
  Line() {}
  Line(pair<int, int> itv, long long val) : itv(itv), val(val) {}
  bool operator<(const Line& line) const { return itv < line.itv; }
};
bool cmp_val(const Line& l1, const Line& l2) { return l1.val < l2.val; }
set<Line> st;
set<Line>::iterator it, it1, it2;
void oper1(int l, int r, long long x) {
  it1 = st.upper_bound(Line(pair<int, int>(l, inf), 0ll));
  --it1;
  it2 = st.upper_bound(Line(pair<int, int>(r, inf), 0ll));
  vector<Line> buf;
  for (it = it1; it != it2; ++it) {
    if ((it->itv).first > r) break;
    buf.push_back(*it);
  }
  st.erase(it1, it2);
  if (buf[0].itv.first < l) {
    st.insert(Line(pair<int, int>(buf[0].itv.first, l - 1), buf[0].val));
    buf[0].itv.first = l;
  }
  int sz = buf.size();
  if (buf[sz - 1].itv.second > r) {
    st.insert(
        Line(pair<int, int>(r + 1, buf[sz - 1].itv.second), buf[sz - 1].val));
    buf[sz - 1].itv.second = r;
  }
  for (Line& line : buf) {
    line.val += x;
    st.insert(line);
  }
}
void oper2(int l, int r, long long x) {
  it1 = st.upper_bound(Line(pair<int, int>(l, inf), 0ll));
  --it1;
  it2 = st.upper_bound(Line(pair<int, int>(r, inf), 0ll));
  vector<Line> buf;
  for (it = it1; it != it2; ++it) {
    if ((it->itv).first > r) break;
    buf.push_back(*it);
  }
  st.erase(it1, it2);
  if (buf[0].itv.first < l && buf[0].val != x) {
    st.insert(Line(pair<int, int>(buf[0].itv.first, l - 1), buf[0].val));
    buf[0].itv.first = l;
  }
  int sz = buf.size();
  if (buf[sz - 1].itv.second > r && buf[sz - 1].val != x) {
    st.insert(
        Line(pair<int, int>(r + 1, buf[sz - 1].itv.second), buf[sz - 1].val));
    buf[sz - 1].itv.second = r;
  }
  st.insert(Line(pair<int, int>(buf[0].itv.first, buf[sz - 1].itv.second), x));
}
long long oper3(int l, int r, long long x) {
  it1 = st.upper_bound(Line(pair<int, int>(l, inf), 0ll));
  --it1;
  it2 = st.upper_bound(Line(pair<int, int>(r, inf), 0ll));
  vector<Line> buf;
  for (it = it1; it != it2; ++it) {
    if ((it->itv).first > r) break;
    buf.push_back(*it);
  }
  if (buf[0].itv.first < l) {
    buf[0].itv.first = l;
  }
  int sz = buf.size();
  if (buf[sz - 1].itv.second > r) {
    buf[sz - 1].itv.second = r;
  }
  sort(buf.begin(), buf.end(), cmp_val);
  long long w = 0;
  for (Line& line : buf) {
    w += (line.itv.second - line.itv.first + 1);
    if (w >= x) return line.val;
  }
  return -1;
}
long long qpow(long long a, long long b, long long mod) {
  long long ret = 1;
  a = a % mod;
  while (b > 0) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long oper4(int l, int r, long long x, long long y) {
  it1 = st.upper_bound(Line(pair<int, int>(l, inf), 0ll));
  --it1;
  it2 = st.upper_bound(Line(pair<int, int>(r, inf), 0ll));
  vector<Line> buf;
  for (it = it1; it != it2; ++it) {
    if ((it->itv).first > r) break;
    buf.push_back(*it);
  }
  if (buf[0].itv.first < l) {
    buf[0].itv.first = l;
  }
  int sz = buf.size();
  if (buf[sz - 1].itv.second > r) {
    buf[sz - 1].itv.second = r;
  }
  long long ret = 0;
  for (Line& line : buf) {
    int w = (line.itv.second - line.itv.first + 1);
    ret += (long long)w * qpow(line.val, x, y) % y;
    ret %= y;
  }
  return ret;
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 1; i <= n; ++i) a[i] = (rnd() % vmax) + 1;
  st.clear();
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j + 1 <= n && a[i] == a[j + 1]) ++j;
    st.insert(Line(pair<int, int>(i, j), a[i]));
    i = j;
  }
  int op, l, r;
  long long x, y;
  int cnt = 0;
  for (int i = 1; i <= m; ++i) {
    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;
    if (op == 1)
      oper1(l, r, x);
    else if (op == 2)
      oper2(l, r, x);
    else if (op == 3) {
      long long ret = oper3(l, r, x);
      cout << ret << endl;
    } else {
      long long ret = oper4(l, r, x, y);
      cout << ret << endl;
    }
  }
  return 0;
}