Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double PI = 3.141592653589793;
template <typename T>
class segtree {
 private:
  int N;
  T lambda(T a, T b) { return {a.first + b.first, max(a.second, b.second)}; }
  T rseg(int l, int r, int i, int lt, int rt) {
    if (l == lt && r == rt)
      return tree[i];
    else {
      int m = (lt + rt) >> 1;
      if (m < r && m >= l)
        return lambda(rseg(l, m, i << 1, lt, m),
                      rseg(m + 1, r, ((i << 1) | 1), m + 1, rt));
      else if (r <= m)
        return rseg(l, r, i << 1, lt, m);
      else if (l > m)
        return rseg(l, r, ((i << 1) | 1), m + 1, rt);
      else
        return {};
    }
  }
  void upseg(int index, T val, int i, int l, int r) {
    if (l == r)
      tree[i] = val;
    else {
      int m = (l + r) >> 1;
      if (index <= m)
        upseg(index, val, i << 1, l, m);
      else if (index > m)
        upseg(index, val, ((i << 1) | 1), m + 1, r);
      tree[i] = lambda(tree[i << 1], tree[((i << 1) | 1)]);
    }
  }
  void preseg(int i, int l, int r, const vector<T> &in) {
    if (l == r)
      tree[i] = in[l];
    else {
      int m = (l + r) >> 1;
      preseg(i << 1, l, m, in);
      preseg(((i << 1) | 1), m + 1, r, in);
      tree[i] = lambda(tree[i << 1], tree[((i << 1) | 1)]);
    }
  }

 public:
  vector<T> tree;
  segtree(int siz) {
    N = siz;
    tree.resize(4 * N);
  }
  segtree(const vector<T> &in) {
    N = in.size();
    tree.resize(4 * N);
    preseg(1, 0, N - 1, in);
  }
  T rng(int l, int r) { return rseg(l, r, 1, 0, N - 1); }
  void upd(int ind, T val) { upseg(ind, val, 1, 0, N - 1); }
};
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<long long, pair<int, int>>> in(n);
  for (int i = 0; i < n; i++) {
    cin >> in[i].first;
    in[i].second.first = in[i].first;
    in[i].second.second = i;
  }
  segtree<pair<long long, pair<int, int>>> tree(in);
  while (m--) {
    int t;
    cin >> t;
    if (t == 1) {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      auto ans = tree.rng(l, r);
      cout << ans.first << '\n';
    } else if (t == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      l--;
      r--;
      auto proc = tree.rng(l, r);
      while (proc.second.first >= x) {
        tree.upd(proc.second.second,
                 {proc.second.first % x,
                  {proc.second.first % x, proc.second.second}});
        proc = tree.rng(l, r);
      }
    } else {
      int k, x;
      cin >> k >> x;
      k--;
      tree.upd(k, {x, {x, k}});
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  for (int c = 1; c < T + 1; c++) {
    solve();
  }
}