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 = 100005;
long long sumv[maxn << 2], maxv[maxn << 2];
int n, q;
void push_up(int rt) {
  sumv[rt] = sumv[rt << 1] + sumv[rt << 1 | 1];
  maxv[rt] = max(maxv[rt << 1], maxv[rt << 1 | 1]);
}
void build(int l, int r, int rt) {
  if (l == r) {
    int num;
    cin >> num;
    sumv[rt] = num;
    maxv[rt] = num;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  push_up(rt);
}
void update1(int p, int val, int l, int r, int rt) {
  if (l == r) {
    sumv[rt] = val;
    maxv[rt] = val;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update1(p, val, l, m, rt << 1);
  else
    update1(p, val, m + 1, r, rt << 1 | 1);
  push_up(rt);
}
void update2(int cl, int cr, int val, int l, int r, int rt) {
  if (maxv[rt] < val) return;
  if (l == r) {
    sumv[rt] %= val;
    maxv[rt] %= val;
    return;
  }
  int m = (l + r) >> 1;
  if (cl <= m) update2(cl, cr, val, l, m, rt << 1);
  if (cr > m) update2(cl, cr, val, m + 1, r, rt << 1 | 1);
  push_up(rt);
}
long long query(int ql, int qr, int l, int r, int rt) {
  if (ql <= l && r <= qr) return sumv[rt];
  int m = (l + r) >> 1;
  long long sumr = 0;
  if (ql <= m) sumr += query(ql, qr, l, m, rt << 1);
  if (qr > m) sumr += query(ql, qr, m + 1, r, rt << 1 | 1);
  return sumr;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  build(1, n, 1);
  int type, l, r, x;
  while (q--) {
    cin >> type >> l >> r;
    if (1 == type)
      cout << query(l, r, 1, n, 1) << endl;
    else if (2 == type) {
      cin >> x;
      update2(l, r, x, 1, n, 1);
    } else
      update1(l, r, 1, n, 1);
  }
  return 0;
}