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 long long N = 2e5 + 9;
long long a[N], n, m;
struct node {
  long long l, r, v, mx;
} tr[N << 2];
void up(long long o) {
  tr[o].v = tr[(o * 2)].v + tr[(o * 2 + 1)].v;
  tr[o].mx = max(tr[(o * 2)].mx, tr[(o * 2 + 1)].mx);
}
void build(long long o, long long l, long long r) {
  tr[o].l = l;
  tr[o].r = r;
  if (l == r) {
    tr[o].v = tr[o].mx = a[l];
    return;
  }
  build((o * 2), l, ((l + r) >> 1));
  build((o * 2 + 1), ((l + r) >> 1) + 1, r);
  up(o);
}
void modi(long long o, long long x, long long y, long long p) {
  long long l = tr[o].l, r = tr[o].r;
  if (tr[o].mx < p) return;
  if (l == r) {
    tr[o].v = tr[o].mx = tr[o].v % p;
    return;
  }
  if (x <= ((l + r) >> 1)) modi((o * 2), x, y, p);
  if (y > ((l + r) >> 1)) modi((o * 2 + 1), x, y, p);
  up(o);
}
void simo(long long o, long long p, long long x) {
  long long l = tr[o].l, r = tr[o].r;
  if (l == r) {
    tr[o].v = tr[o].mx = x;
    return;
  }
  if (p <= ((l + r) >> 1)) simo((o * 2), p, x);
  if (p > ((l + r) >> 1)) simo((o * 2 + 1), p, x);
  up(o);
}
long long query(long long o, long long x, long long y) {
  long long l = tr[o].l, r = tr[o].r, res = 0;
  if (l >= x && r <= y) return tr[o].v;
  if (x <= ((l + r) >> 1)) res += query((o * 2), x, y);
  if (y > ((l + r) >> 1)) res += query((o * 2 + 1), x, y);
  return res;
}
signed main() {
  scanf("%lld %lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build(1, 1, n);
  for (long long i = 1; i <= m; i++) {
    long long opt, l, r, x;
    scanf("%lld %lld %lld", &opt, &l, &r);
    if (opt == 1) cout << query(1, l, r) << "\n";
    if (opt == 2) scanf("%lld", &x), modi(1, l, r, x);
    if (opt == 3) simo(1, l, r);
  }
  return 0;
}