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;
const long long M = 1e9 + 7, maxn = 2e5 + 20;
long long int n, k, m, q, sizev, x, y;
pair<pair<long long int, long long int>, long long int> Max[maxn << 2];
void upd(long long int pos, long long int val, long long int id = 1,
         long long int l = 0, long long int r = n) {
  if (r - l == 1) {
    Max[id] = make_pair(make_pair(val, l), val);
    return;
  }
  long long int mid = (l + r) >> 1;
  if (pos < mid)
    upd(pos, val, id << 1, l, mid);
  else
    upd(pos, val, id << 1 | 1, mid, r);
  pair<pair<long long int, long long int>, long long int> chl = Max[id << 1];
  pair<pair<long long int, long long int>, long long int> chr =
      Max[id << 1 | 1];
  if (chl.first.first > chr.first.first)
    Max[id] = make_pair(make_pair(chl.first.first, chl.first.second),
                        chl.second + chr.second);
  else
    Max[id] = make_pair(make_pair(chr.first.first, chr.first.second),
                        chl.second + chr.second);
}
pair<pair<long long int, long long int>, long long int> get(
    long long int s, long long int e, long long int id = 1, long long int l = 0,
    long long int r = n) {
  if (s >= r || l >= e) return make_pair(make_pair(-1, -1), 0);
  if (s <= l && r <= e) return Max[id];
  long long int mid = (l + r) >> 1;
  pair<pair<long long int, long long int>, long long int> chl =
      get(s, e, id << 1, l, mid);
  pair<pair<long long int, long long int>, long long int> chr =
      get(s, e, id << 1 | 1, mid, r);
  if (chl.first.first > chr.first.first)
    return make_pair(make_pair(chl.first.first, chl.first.second),
                     chl.second + chr.second);
  else
    return make_pair(make_pair(chr.first.first, chr.first.second),
                     chl.second + chr.second);
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    upd(i, a[i]);
  }
  for (long long int i = 0; i < q; i++) {
    cin >> k;
    if (k == 1) {
      cin >> x >> y;
      pair<pair<long long int, long long int>, long long int> pans =
          get(x - 1, y);
      cout << pans.second << endl;
    } else if (k == 2) {
      cin >> x >> y >> k;
      pair<pair<long long int, long long int>, long long int> mmax =
          get(x - 1, y);
      while (mmax.first.first >= k) {
        upd(mmax.first.second, mmax.first.first % k);
        mmax = get(x - 1, y);
      }
    } else {
      cin >> x >> y;
      upd(x - 1, y);
    }
  }
  return 0;
}