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;
ostream &operator<<(ostream &os, vector<long long> &x) {
  for (long long i = 0; i < x.size(); i++) os << x[i] << ' ';
  return os;
}
ostream &operator<<(ostream &os, pair<long long, long long> x) {
  cout << x.first << ' ' << x.second << ' ';
  return os;
}
ostream &operator<<(ostream &os, vector<pair<long long, long long> > &x) {
  for (long long i = 0; i < x.size(); i++) os << x[i] << '\n';
  return os;
}
long long n, q, seed, vmax;
long long a[100005];
set<pair<long long, long long> > st;
const long long mod = 1e9 + 7;
long long rnd() {
  long long ret = seed;
  seed = (seed * 7 + 13) % mod;
  return ret;
}
void makeset() {
  for (long long i = 1; i <= n; i++) st.insert(make_pair(i, a[i]));
  st.insert(make_pair(n + 1, 1e18 + 1));
}
set<pair<long long, long long> >::iterator findFirstQuery(long long l) {
  auto it = st.upper_bound(make_pair(l, 1e18));
  it--;
  return it;
}
set<pair<long long, long long> >::iterator findLastQuery(long long l) {
  auto it = st.upper_bound(make_pair(l, 1e18));
  it--;
  return it;
}
void firstQuery(long long l, long long r, long long x) {
  auto left = findFirstQuery(l);
  auto right = findLastQuery(r);
  if (left == right) {
    auto it = left->second;
    if (left->first == l) st.erase(left);
    st.insert(make_pair(l, it + x));
    right = findLastQuery(r + 1);
    if (r + 1 != right->first) st.insert(make_pair(r + 1, it));
    return;
  }
  left++;
  vector<pair<long long, long long> > newQuery;
  for (auto it = left; it != right; it++) newQuery.push_back(*it);
  for (auto &val : newQuery) {
    st.erase(st.find(val));
    val.second += x;
  }
  right--;
  for (auto val : newQuery) st.insert(val);
  left = findFirstQuery(l);
  pair<long long, long long> rec = *left;
  if (rec.first == l) st.erase(left);
  rec.first = l;
  rec.second += x;
  st.insert(rec);
  right = findLastQuery(r);
  rec = *right;
  st.erase(right);
  st.insert(make_pair(rec.first, rec.second + x));
  right = findLastQuery(r + 1);
  if (r + 1 != right->first) st.insert(make_pair(r + 1, rec.second));
}
void secQuery(long long l, long long r, long long x, long long zzz) {
  auto left = findFirstQuery(l);
  auto right = findLastQuery(r);
  if (left == right) {
    auto it = left->second;
    if (left->first == l) st.erase(left);
    st.insert(make_pair(l, x));
    right = findLastQuery(r + 1);
    if (r + 1 != right->first) st.insert(make_pair(r + 1, it));
    return;
  }
  left++;
  vector<pair<long long, long long> > newQuery;
  for (auto it = left; it != right; it++) newQuery.push_back(*it);
  for (auto &val : newQuery) {
    st.erase(st.find(val));
  }
  right--;
  left = findFirstQuery(l);
  pair<long long, long long> rec = *left;
  if (rec.first == l) st.erase(left);
  rec.first = l;
  rec.second = x;
  st.insert(rec);
  right = findLastQuery(r);
  rec = *right;
  st.erase(right);
  st.insert(make_pair(rec.first, x));
  right = findLastQuery(r + 1);
  if (r + 1 != right->first) st.insert(make_pair(r + 1, rec.second));
}
void thirdQuery(long long l, long long r, long long k, long long zzz) {
  vector<pair<long long, long long> > tmp;
  auto left = findFirstQuery(l);
  auto right = findLastQuery(r);
  if (left == right) {
    cout << left->second << '\n';
    return;
  }
  left++;
  for (auto it = left; it != right; it++) {
    auto it2 = it;
    it2++;
    tmp.push_back(make_pair(it->second, it2->first - it->first));
  }
  right--;
  long long len = left->first - l;
  left--;
  tmp.push_back(make_pair(left->second, len));
  right++;
  assert(r - right->first + 1 > 0);
  len = r - right->first + 1;
  tmp.push_back(make_pair(right->second, len));
  sort(tmp.begin(), tmp.end());
  long long cnt = 0;
  for (long long i = 0; i < tmp.size(); i++) {
    cnt += tmp[i].second;
    if (cnt >= k) {
      cout << tmp[i].first << '\n';
      return;
    }
  }
}
long long binPow(long long a, long long q, long long modx) {
  long long ret = 1;
  a %= modx;
  for (long long i = 0; (1LL << i) <= q; i++) {
    if (q & (1LL << i)) ret = (ret * a) % modx;
    a = (a * a) % modx;
  }
  return ret;
}
void fourthQuery(long long l, long long r, long long x, long long y,
                 long long zzz) {
  long long ret = 0;
  auto left = findFirstQuery(l);
  auto right = findLastQuery(r);
  if (left == right) {
    ret = (r - l + 1) * binPow(left->second, x, y) % y;
    cout << ret << '\n';
    return;
  }
  left++;
  for (auto it = left; it != right; it++) {
    auto it2 = it;
    it2++;
    ret = (ret + (it2->first - it->first) * binPow(it->second, x, y) % y) % y;
  }
  long long len = left->first - l;
  left--;
  ret = (ret + len * binPow(left->second, x, y) % y) % y;
  len = r - right->first + 1;
  ret = (ret + len * binPow(right->second, x, y) % y) % y;
  cout << ret << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q >> seed >> vmax;
  for (long long i = 1; i <= n; i++) a[i] = rnd() % vmax + 1;
  makeset();
  for (long long i = 1; i <= q; i++) {
    long long op = (rnd() % 4) + 1;
    long long l = (rnd() % n) + 1;
    long long r = (rnd() % n) + 1;
    if (l > r) swap(l, r);
    long long x = 0, y = 0;
    if (op == 3)
      x = (rnd() % (r - l + 1)) + 1;
    else
      x = (rnd() % vmax) + 1;
    if (op == 4) y = (rnd() % vmax) + 1;
    if (op == 1) {
      firstQuery(l, r, x);
    } else if (op == 2) {
      secQuery(l, r, x, i);
    } else if (op == 3) {
      thirdQuery(l, r, x, i);
    } else {
      fourthQuery(l, r, x, y, i);
    }
  }
}