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;
int md = 1e9 + 7;
int n, m, vmax;
long long seed;
int op, l, r, x, y;
int rnd() {
  int ans = (int)seed;
  seed = (seed * 7 + 13) % (long long)md;
  return ans;
}
void params() {
  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 pow(long long a, int b, long long m) {
  a %= m;
  if (b == 0) return 1;
  if (b == 1) return a;
  long long ans = pow(a, b / 2, m);
  ans = (ans * ans) % m;
  if (b % 2) ans = (ans * a) % m;
  return ans;
}
const int mx = 1 << 17;
int uni[2 * mx], pflag[2 * mx];
long long tree[2 * mx], del[2 * mx];
void init() {
  for (int i = mx; i < 2 * mx; i++) uni[i] = 1;
  int i = mx;
  while (--i) {
    uni[i] =
        uni[i << 1] && uni[i << 1 | 1] && (tree[i << 1] == tree[i << 1 | 1]);
    if (uni[i]) tree[i] = tree[i << 1];
  }
}
vector<pair<long long, int>> ranges;
void push_update(int i) {
  if (pflag[i] == 1) {
    if (uni[i]) {
      tree[i << 1] = tree[i << 1 | 1] = tree[i];
      del[i << 1] = del[i << 1 | 1] = 0;
      uni[i << 1] = uni[i << 1 | 1] = 1;
      pflag[i << 1] = pflag[i << 1 | 1] = 1;
    } else {
      if (uni[i << 1])
        tree[i << 1] += del[i];
      else
        del[i << 1] += del[i];
      if (uni[i << 1 | 1])
        tree[i << 1 | 1] += del[i];
      else
        del[i << 1 | 1] += del[i];
      pflag[i << 1] = pflag[i << 1 | 1] = 1;
      del[i] = 0;
    }
  }
  pflag[i] = 0;
}
long long query(int i, int l, int r, int u, int v) {
  int mid = (l + r) / 2;
  if (u <= l && r <= v) {
    if (op == 1) {
      if (uni[i])
        tree[i] += x;
      else
        del[i] += x;
      pflag[i] = 1;
    } else if (op == 2) {
      tree[i] = x;
      del[i] = 0;
      uni[i] = 1;
      pflag[i] = 1;
    } else if (op == 3) {
      if (uni[i])
        ranges.push_back(make_pair(tree[i], r - l));
      else {
        push_update(i);
        query(i << 1, l, mid, l, mid);
        query(i << 1 | 1, mid, r, mid, r);
      }
    } else {
      if (uni[i]) {
        return (pow(tree[i], x, y) * (r - l)) % (long long)y;
      } else {
        push_update(i);
        long long ans = query(i << 1, l, mid, l, mid);
        ans += query(i << 1 | 1, mid, r, mid, r);
        ans %= (long long)y;
        return ans;
      }
    }
    return 0;
  }
  push_update(i);
  long long ans = 0;
  if (u < mid) ans += query(i << 1, l, mid, u, v);
  if (mid < v) ans += query(i << 1 | 1, mid, r, u, v);
  if (op == 4) ans %= (long long)y;
  uni[i] = uni[i << 1] && uni[i << 1 | 1] && (tree[i << 1] == tree[i << 1 | 1]);
  if (uni[i]) tree[i] = tree[i << 1];
  return ans;
}
int main() {
  cin >> n >> m >> seed >> vmax;
  for (int i = 0; i < n; i++) {
    tree[mx + i] = (rnd() % vmax) + 1;
  }
  init();
  for (int i = 1; i <= m; i++) {
    ranges.clear();
    params();
    long long ans = query(1, mx, 2 * mx, mx + l - 1, mx + r);
    if (op == 3) {
      sort((ranges).begin(), (ranges).end());
      int cnt = 0;
      for (auto rng : ranges) {
        cnt += rng.second;
        if (x <= cnt) {
          ans = rng.first;
          break;
        }
      }
    }
    if (op >= 3)
      printf(
          "%lld"
          "\n",
          ans);
  }
  return 0;
}