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;
int n, m, a[100100];
bool Compare(int x, int y) { return a[x] < a[y]; }
struct Segtree {
  Segtree *ls, *rs;
  int max_pos;
  long long sum;
  void* operator new(size_t) {
    static Segtree mempool[100100 << 1], *C = mempool;
    return C++;
  }
  void Push_Up() {
    max_pos = max(ls->max_pos, rs->max_pos, Compare);
    sum = ls->sum + rs->sum;
  }
  void Build_Tree(int x, int y) {
    int mid = x + y >> 1;
    if (x == y) {
      max_pos = mid;
      sum = a[mid];
      return;
    }
    (ls = new Segtree)->Build_Tree(x, mid);
    (rs = new Segtree)->Build_Tree(mid + 1, y);
    Push_Up();
  }
  void Modify(int x, int y, int pos) {
    int mid = x + y >> 1;
    if (x == y) {
      sum = a[mid];
      return;
    }
    if (pos <= mid)
      ls->Modify(x, mid, pos);
    else
      rs->Modify(mid + 1, y, pos);
    Push_Up();
  }
  int Get_Pos(int x, int y, int l, int r) {
    int mid = x + y >> 1;
    if (x == l && y == r) return max_pos;
    if (r <= mid) return ls->Get_Pos(x, mid, l, r);
    if (l > mid) return rs->Get_Pos(mid + 1, y, l, r);
    return max(ls->Get_Pos(x, mid, l, mid), rs->Get_Pos(mid + 1, y, mid + 1, r),
               Compare);
  }
  long long Get_Sum(int x, int y, int l, int r) {
    int mid = x + y >> 1;
    if (x == l && y == r) return sum;
    if (r <= mid) return ls->Get_Sum(x, mid, l, r);
    if (l > mid) return rs->Get_Sum(mid + 1, y, l, r);
    return ls->Get_Sum(x, mid, l, mid) + rs->Get_Sum(mid + 1, y, mid + 1, r);
  }
}* tree = new Segtree;
int main() {
  int i, p, x, y, z;
  cin >> n >> m;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  tree->Build_Tree(1, n);
  for (i = 1; i <= m; i++) {
    scanf("%d", &p);
    if (p == 1) {
      scanf("%d%d", &x, &y);
      printf("%I64d\n", tree->Get_Sum(1, n, x, y));
    } else if (p == 2) {
      scanf("%d%d%d", &x, &y, &z);
      while (1) {
        int pos = tree->Get_Pos(1, n, x, y);
        if (a[pos] < z) break;
        a[pos] %= z;
        tree->Modify(1, n, pos);
      }
    } else {
      scanf("%d%d", &x, &y);
      a[x] = y;
      tree->Modify(1, n, x);
    }
  }
  return 0;
}