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;
struct node {
  long long sum, rmq;
} st[(long long)1e6];
long long n, m, q;
void build(long long l = 0, long long r = n - 1, long long curr = 1) {
  if (l == r) {
    scanf("%lld", &st[curr].sum);
    st[curr].rmq = st[curr].sum;
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, 2 * curr);
  build(mid + 1, r, 2 * curr + 1);
  st[curr].sum = st[2 * curr].sum + st[2 * curr + 1].sum;
  st[curr].rmq = max(st[2 * curr].rmq, st[2 * curr + 1].rmq);
}
long long sum(long long x, long long y, long long l = 0, long long r = n - 1,
              long long curr = 1) {
  long long mid = (l + r) / 2;
  if (l == x && r == y) {
    return st[curr].sum;
  }
  if (y <= mid)
    return sum(x, y, l, mid, 2 * curr);
  else if (x > mid)
    return sum(x, y, mid + 1, r, 2 * curr + 1);
  return sum(x, mid, l, mid, 2 * curr) +
         sum(mid + 1, y, mid + 1, r, 2 * curr + 1);
}
void set_v(long long x, long long v, long long l = 0, long long r = n - 1,
           long long curr = 1) {
  long long mid = (l + r) / 2;
  if (l == r && l == x) {
    st[curr].sum = v;
    st[curr].rmq = v;
    return;
  }
  if (x <= mid)
    set_v(x, v, l, mid, 2 * curr);
  else
    set_v(x, v, mid + 1, r, 2 * curr + 1);
  st[curr].sum = st[2 * curr].sum + st[2 * curr + 1].sum;
  st[curr].rmq = max(st[2 * curr].rmq, st[2 * curr + 1].rmq);
}
void set_range(long long x, long long y, long long mod, long long l = 0,
               long long r = n - 1, long long curr = 1) {
  long long mid = (l + r) / 2;
  if (l == r) {
    st[curr].sum %= mod;
    st[curr].rmq %= mod;
    return;
  }
  if (l == x && r == y) {
    if (st[2 * curr].rmq >= mod) set_range(x, mid, mod, l, mid, 2 * curr);
    if (st[2 * curr + 1].rmq >= mod)
      set_range(mid + 1, y, mod, mid + 1, r, 2 * curr + 1);
    st[curr].sum = st[2 * curr].sum + st[2 * curr + 1].sum;
    st[curr].rmq = max(st[2 * curr].rmq, st[2 * curr + 1].rmq);
    return;
  }
  if (y <= mid)
    set_range(x, y, mod, l, mid, 2 * curr);
  else if (x > mid)
    set_range(x, y, mod, mid + 1, r, 2 * curr + 1);
  else {
    set_range(x, mid, mod, l, mid, 2 * curr);
    set_range(mid + 1, y, mod, mid + 1, r, 2 * curr + 1);
  }
  st[curr].sum = st[2 * curr].sum + st[2 * curr + 1].sum;
  st[curr].rmq = max(st[2 * curr].rmq, st[2 * curr + 1].rmq);
}
int32_t main() {
  scanf("%lld%lld", &n, &m);
  build();
  while (m--) {
    long long tp;
    scanf("%lld", &tp);
    if (tp == 1) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      printf("%lld\n", sum(l - 1, r - 1));
    } else if (tp == 2) {
      long long l, r, x;
      scanf("%lld%lld%lld", &l, &r, &x);
      set_range(l - 1, r - 1, x);
    } else {
      long long k, x;
      scanf("%lld%lld", &k, &x);
      set_v(k - 1, x);
    }
  }
}