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 INFll = 1000000000000000000ll, INF = 0x3f3f3f3f,
                _MOD = (1000 * 1000 * 1000 + 7);
const long double EPS = 1e-8;
const long long MAXN = 2 * 100000 + 10;
long long n, m, ans, arr[MAXN];
pair<long long, long long> lr[2 * MAXN], nod[2 * MAXN];
void make_tree() {
  for (long long i = 0; i < n; i++)
    lr[n + i] = pair<long long, long long>(i, i + 1),
           nod[n + i] = pair<long long, long long>(arr[i], arr[i]);
  for (long long i = n - 1; i >= 0; i--) {
    lr[i] = pair<long long, long long>(lr[(i << 1)].first,
                                       lr[((i << 1) + 1)].second);
    nod[i] = pair<long long, long long>(
        max(nod[(i << 1)].first, nod[((i << 1) + 1)].first),
        nod[(i << 1)].second + nod[((i << 1) + 1)].second);
  }
}
void add(long long pos, long long l, long long r, long long x) {
  long long lpos = lr[pos].first, rpos = lr[pos].second;
  if (lpos >= r || rpos <= l) return;
  if (rpos - 1 == lpos) {
    long long tmp = nod[pos].second % x;
    nod[pos] = pair<long long, long long>(tmp, tmp);
    return;
  }
  if (nod[pos].first < x) return;
  add((pos << 1), l, r, x), add(((pos << 1) + 1), l, r, x);
  nod[pos] = pair<long long, long long>(
      max(nod[(pos << 1)].first, nod[((pos << 1) + 1)].first),
      nod[(pos << 1)].second + nod[((pos << 1) + 1)].second);
}
void get(long long pos, long long l, long long r) {
  long long lpos = lr[pos].first, rpos = lr[pos].second;
  if (lpos >= r || rpos <= l) return;
  if (lpos >= l && rpos <= r) {
    ans += nod[pos].second;
    return;
  }
  get((pos << 1), l, r), get(((pos << 1) + 1), l, r);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> arr[i];
  n = pow(2, ceil(log2(n)));
  make_tree();
  for (long long i = 0; i < m; i++) {
    long long c;
    cin >> c;
    switch (c) {
      long long l, r, x, k;
      case 2:
        cin >> l >> r >> x;
        l--;
        add(1, l, r, x);
        continue;
      case 1:
        cin >> l >> r;
        l--;
        get(1, l, r);
        cout << ans << "\n";
        ans = 0;
        continue;
      case 3:
        cin >> k >> x;
        k--;
        long long pos = k + n;
        nod[pos] = pair<long long, long long>(x, x);
        pos = (pos >> 1);
        while (pos) {
          nod[pos] = pair<long long, long long>(
              max(nod[(pos << 1)].first, nod[((pos << 1) + 1)].first),
              nod[(pos << 1)].second + nod[((pos << 1) + 1)].second);
          pos = (pos >> 1);
        }
    }
  }
}