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 int mx = 1e5 + 10;
static pair<long long, long long> st[5 * mx];
int n, m, ch, qs, qe, k, mod, diff;
static int a[mx], idx[mx];
void build(int x, int y, int n) {
  if (x == y) {
    st[n].second = st[n].first = a[x];
    idx[x] = n;
    return;
  }
  int mid = (x + y) >> 1;
  build(x, mid, 2 * n);
  build(mid + 1, y, 2 * n + 1);
  st[n].first = max(st[2 * n].first, st[2 * n + 1].first);
  st[n].second = st[2 * n].second + st[2 * n + 1].second;
}
void up1(int n) {
  st[n].first = st[n].second = a[k];
  n /= 2;
  while (n) {
    st[n].first = max(st[2 * n].first, st[2 * n + 1].first);
    st[n].second = st[2 * n].second + st[2 * n + 1].second;
    n /= 2;
  }
}
long long qu(int x, int y, int n) {
  if (y < qs || qe < x) return 0;
  if (qs <= x && y <= qe) return st[n].second;
  int mid = (x + y) >> 1;
  return qu(x, mid, 2 * n) + qu(mid + 1, y, 2 * n + 1);
}
void propagate(int x, int y, int n) {
  if (x == y) {
    a[x] %= mod;
    st[n].second = st[n].first = a[x];
    return;
  }
  int mid = (x + y) >> 1;
  if (st[2 * n].first >= mod) propagate(x, mid, 2 * n);
  if (st[2 * n + 1].first >= mod) propagate(mid + 1, y, 2 * n + 1);
  st[n].first = max(st[2 * n].first, st[2 * n + 1].first);
  st[n].second = st[2 * n].second + st[2 * n + 1].second;
}
void up2(int x, int y, int n) {
  if (y < qs || qe < x) return;
  if (qs <= x && y <= qe) {
    propagate(x, y, n);
    return;
  }
  int mid = (x + y) >> 1;
  up2(x, mid, 2 * n);
  up2(mid + 1, y, 2 * n + 1);
  st[n].first = max(st[2 * n].first, st[2 * n + 1].first);
  st[n].second = st[2 * n].second + st[2 * n + 1].second;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build(1, n, 1);
  while (m--) {
    cin >> ch;
    if (ch == 1) {
      cin >> qs >> qe;
      cout << qu(1, n, 1) << '\n';
    } else if (ch == 2) {
      cin >> qs >> qe >> mod;
      up2(1, n, 1);
    } else {
      cin >> k >> mod;
      a[k] = mod;
      up1(idx[k]);
    }
  }
  return 0;
}