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 mk, n, m, base[200010];
int val[200010 << 1];
long long s[200010 << 1];
int Max(int a, int b) { return a > b ? a : b; }
void build(int rt, int l, int r) {
  if (l == r) return val[rt] = s[rt] = base[l], void();
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
  val[rt] = Max(val[rt << 1], val[rt << 1 | 1]);
}
void _change(int rt, int l, int r, int p, int v) {
  if (l == r) return val[rt] = s[rt] = v, void();
  int mid = (l + r) >> 1;
  if (p <= mid)
    _change(rt << 1, l, mid, p, v);
  else
    _change(rt << 1 | 1, mid + 1, r, p, v);
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
  val[rt] = Max(val[rt << 1], val[rt << 1 | 1]);
}
void update(int rt, int l, int r, int ul, int ur, int mod) {
  if (l == r) return val[rt] %= mod, s[rt] %= mod, void();
  int mid = (l + r) >> 1;
  if (ul <= mid && val[rt << 1] >= mod) update(rt << 1, l, mid, ul, ur, mod);
  if (ur > mid && val[rt << 1 | 1] >= mod)
    update(rt << 1 | 1, mid + 1, r, ul, ur, mod);
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
  val[rt] = Max(val[rt << 1], val[rt << 1 | 1]);
}
long long query(int rt, int l, int r, int ql, int qr) {
  int mid = (l + r) >> 1;
  long long res = 0;
  if (l >= ql && r <= qr) return s[rt];
  if (ql <= mid) res += query(rt << 1, l, mid, ql, qr);
  if (qr > mid) res += query(rt << 1 | 1, mid + 1, r, ql, qr);
  return res;
}
int main() {
  cin >> n >> m;
  int l, r, v;
  for (int i = 1; i <= n; ++i) scanf("%d", &base[i]);
  build(1, 1, n);
  while (m--) {
    scanf("%d", &mk);
    if (mk == 1)
      scanf("%d%d", &l, &r), printf("%lld\n", query(1, 1, n, l, r));
    else if (mk == 2)
      scanf("%d%d%d", &l, &r, &v), update(1, 1, n, l, r, v);
    else
      scanf("%d%d", &r, &v), _change(1, 1, n, r, v);
  }
}