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 MAXn = 1e5 + 10, MAXx = 4 * MAXn;
struct salar {
  int l, r, max1, leftchild, rightchild, id;
  long long sum;
  salar() { sum = id = l = r = max1 = leftchild = rightchild = 0; }
  void idit(int v) {
    id = v;
    leftchild = 2 * v;
    rightchild = 2 * v + 1;
  }
};
salar seg[MAXx];
int ar[MAXn], n, m;
void build(int v, int l, int r) {
  seg[v].l = l, seg[v].r = r;
  seg[v].idit(v);
  if (r - l == 1) {
    seg[v].sum = ar[l];
    seg[v].max1 = ar[l];
    return;
  }
  int med = (l + r) / 2;
  build(seg[v].leftchild, l, med);
  build(seg[v].rightchild, med, r);
  seg[v].sum = seg[seg[v].leftchild].sum + seg[seg[v].rightchild].sum;
  seg[v].max1 = max(seg[seg[v].leftchild].max1, seg[seg[v].rightchild].max1);
}
long long get(int v, int l, int r) {
  if (max(l, seg[v].l) >= min(r, seg[v].r)) return 0;
  if (l <= seg[v].l && seg[v].r <= r) {
    return seg[v].sum;
  }
  return get(seg[v].leftchild, l, r) + get(seg[v].rightchild, l, r);
}
void domodule(int v, int l, int r, int x) {
  if (seg[v].max1 < x) return;
  if (max(l, seg[v].l) >= min(r, seg[v].r)) return;
  if (seg[v].r - seg[v].l == 1) {
    ar[seg[v].l] %= x;
    seg[v].sum = ar[seg[v].l];
    seg[v].max1 = ar[seg[v].l];
    return;
  }
  domodule(seg[v].leftchild, l, r, x);
  domodule(seg[v].rightchild, l, r, x);
  seg[v].sum = seg[seg[v].leftchild].sum + seg[seg[v].rightchild].sum;
  seg[v].max1 = max(seg[seg[v].leftchild].max1, seg[seg[v].rightchild].max1);
}
void change(int v, int l, int x) {
  if (l < seg[v].l || l >= seg[v].r) return;
  if (seg[v].r - seg[v].l == 1) {
    ar[seg[v].l] = x;
    seg[v].sum = ar[seg[v].l];
    seg[v].max1 = ar[seg[v].l];
    return;
  }
  change(seg[v].leftchild, l, x);
  change(seg[v].rightchild, l, x);
  seg[v].sum = seg[seg[v].leftchild].sum + seg[seg[v].rightchild].sum;
  seg[v].max1 = max(seg[seg[v].leftchild].max1, seg[seg[v].rightchild].max1);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> ar[i];
  build(1, 0, n);
  for (int i = 0; i < m; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l--;
      cout << get(1, l, r) << endl;
    } else if (type == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      l--;
      domodule(1, l, r, x);
    } else {
      int l, x;
      cin >> l >> x;
      l--;
      change(1, l, x);
    }
  }
}