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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = gc();
  bool positive = 1;
  for (; !isdigit(ch); ch = gc())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = gc()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
struct tree {
  long long l, r, sum, maxn;
} st[100005 * 4];
long long n, m, x, y, t, o, a[100005];
inline void update(long long node) {
  st[node].sum = st[node * 2].sum + st[node * 2 + 1].sum;
  st[node].maxn = max(st[node * 2].maxn, st[node * 2 + 1].maxn);
}
inline void build(long long node, long long l, long long r) {
  st[node].l = l;
  st[node].r = r;
  if (l == r) {
    st[node].sum = st[node].maxn = a[l];
    return;
  }
  build(node * 2, l, (l + r) / 2);
  build(node * 2 + 1, (l + r) / 2 + 1, r);
  update(node);
}
inline void change(long long node, long long l, long long r, long long t) {
  if (st[node].maxn < t) return;
  if (st[node].l > r || st[node].r < l) return;
  if (st[node].l == st[node].r) {
    st[node].sum %= t;
    st[node].maxn %= t;
    return;
  }
  change(node * 2, l, r, t);
  change(node * 2 + 1, l, r, t);
  update(node);
}
inline void change1(long long node, long long id, long long t) {
  if (st[node].l > id || st[node].r < id) return;
  if (st[node].l == st[node].r && st[node].l == id) {
    st[node].sum = st[node].maxn = t;
    return;
  }
  change1(node * 2, id, t);
  change1(node * 2 + 1, id, t);
  update(node);
}
inline long long query(long long node, long long l, long long r) {
  if (st[node].l > r || st[node].r < l) return 0;
  if (st[node].l >= l && st[node].r <= r) {
    return st[node].sum;
  }
  return query(node * 2, l, r) + query(node * 2 + 1, l, r);
}
signed main() {
  n = read();
  m = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  build(1, 1, n);
  while (m--) {
    o = read();
    if (o == 1) {
      x = read();
      y = read();
      writeln(query(1, x, y));
    }
    if (o == 2) {
      x = read();
      y = read();
      t = read();
      change(1, x, y, t);
    }
    if (o == 3) {
      x = read();
      t = read();
      change1(1, x, t);
    }
  }
  return 0;
}