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 inf = 0x3f3f3f3f;
const int mod = 1e8;
const int Max = 1e5 + 10;
struct Node {
  int l, r;
  long long sum;
  long long max_val;
} node[Max << 2];
void push_up(int num) {
  node[num].sum = node[num << 1].sum + node[num << 1 | 1].sum;
  node[num].max_val = max(node[num << 1].max_val, node[num << 1 | 1].max_val);
}
void build(int l, int r, int num) {
  node[num].l = l;
  node[num].r = r;
  if (l == r) {
    scanf("%lld", &node[num].sum);
    node[num].max_val = node[num].sum;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, num << 1);
  build(mid + 1, r, num << 1 | 1);
  push_up(num);
}
void upData(int l, int r, long long x, int num) {
  if (node[num].max_val < x) return;
  if (node[num].l == node[num].r) {
    node[num].max_val = node[num].sum = node[num].sum % x;
    return;
  }
  int mid = (node[num].l + node[num].r) >> 1;
  if (l <= mid) upData(l, r, x, num << 1);
  if (r > mid) upData(l, r, x, num << 1 | 1);
  push_up(num);
}
void upData(int pos, long long val, int num) {
  if (node[num].l == node[num].r) {
    node[num].max_val = node[num].sum = val;
    return;
  }
  int mid = (node[num].l + node[num].r) >> 1;
  if (pos <= mid) upData(pos, val, num << 1);
  if (pos > mid) upData(pos, val, num << 1 | 1);
  push_up(num);
}
long long query(int l, int r, int num) {
  if (l <= node[num].l && node[num].r <= r) return node[num].sum;
  int mid = (node[num].l + node[num].r) >> 1;
  long long ans = 0;
  if (l <= mid) ans += query(l, r, num << 1);
  if (r > mid) ans += query(l, r, num << 1 | 1);
  return ans;
}
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    build(1, n, 1);
    while (m--) {
      int type, a, b;
      long long x;
      scanf("%d%d%d", &type, &a, &b);
      if (type == 1)
        printf("%lld\n", query(a, b, 1));
      else if (type == 2)
        scanf("%lld", &x), upData(a, b, x, 1);
      else
        upData(a, b, 1);
    }
  }
  return 0;
}