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 INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const long long MAXN = 2e5 + 10;
void F() { ios::sync_with_stdio(false); }
long long n, m;
long long arr[MAXN];
struct node {
  long long l, r;
  long long x, mo;
} t[MAXN << 2];
inline void push_up(long long o) {
  t[o].mo = max(t[o << 1].mo, t[o << 1 | 1].mo);
  t[o].x = t[o << 1 | 1].x + t[o << 1].x;
}
inline void build(long long l, long long r, long long o) {
  t[o].l = l, t[o].r = r;
  if (l == r) {
    t[o].x = t[o].mo = arr[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, o << 1);
  build(mid + 1, r, o << 1 | 1);
  push_up(o);
}
inline void update(long long l, long long o, long long x) {
  if (l == t[o].l && t[o].l == t[o].r) {
    t[o].x = t[o].mo = x;
    return;
  }
  long long mid = (t[o].l + t[o].r) >> 1;
  if (l <= mid) update(l, o << 1, x);
  if (l > mid) update(l, o << 1 | 1, x);
  push_up(o);
}
inline void modif(long long l, long long r, long long o, long long p) {
  if (t[o].mo < p) return;
  if (t[o].l == t[o].r) {
    t[o].x = t[o].mo = t[o].mo % p;
    return;
  }
  long long mid = (t[o].l + t[o].r) >> 1;
  if (r <= mid)
    modif(l, r, o << 1, p);
  else if (l > mid)
    modif(l, r, o << 1 | 1, p);
  else {
    modif(l, r, o << 1, p);
    modif(l, r, o << 1 | 1, p);
  }
  push_up(o);
}
inline long long query(long long l, long long r, long long o) {
  if (t[o].l >= l && t[o].r <= r) return t[o].x;
  if (t[o].l > r || t[o].r < l) return 0;
  long long ans = 0;
  long long mid = (t[o].l + t[o].r) >> 1;
  if (r <= mid)
    ans += query(l, r, o << 1);
  else if (l > mid)
    ans += query(l, r, o << 1 | 1);
  else {
    ans += query(l, r, o << 1);
    ans += query(l, r, o << 1 | 1);
  }
  return ans;
}
int main() {
  F();
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) cin >> arr[i];
  build(1, n, 1);
  while (m--) {
    long long op, l, r, x;
    cin >> op;
    ;
    if (op == 1) {
      cin >> l >> r;
      cout << query(l, r, 1) << endl;
    }
    if (op == 2) {
      long long p;
      cin >> l >> r >> p;
      modif(l, r, 1, p);
    }
    if (op == 3) {
      cin >> l >> x;
      update(l, 1, x);
    }
  }
  return 0;
}