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;
using namespace std;
struct node {
  long long left, right, delta;
  long long sum, maxv;
} a[2000000];
long long n, m, c[2000000];
void build(long long p, long long l, long long r) {
  a[p].left = l;
  a[p].right = r;
  if (l == r) {
    a[p].sum = c[l];
    a[p].maxv = c[l];
    return;
  }
  int mid = (l + r) / 2;
  build(2 * p, l, mid);
  build(2 * p + 1, mid + 1, r);
  a[p].sum = a[2 * p].sum + a[2 * p + 1].sum;
  a[p].maxv = max(a[2 * p].maxv, a[2 * p + 1].maxv);
}
void mod(long long p, long long l, long long r, long long d) {
  if (a[p].maxv < d) return;
  if (a[p].left == a[p].right) {
    a[p].sum = a[p].maxv = a[p].sum % d;
    return;
  }
  int mid = (a[p].left + a[p].right) / 2;
  if (r <= mid)
    mod(2 * p, l, r, d);
  else if (l > mid)
    mod(2 * p + 1, l, r, d);
  else {
    mod(2 * p, l, mid, d);
    mod(2 * p + 1, mid + 1, r, d);
  }
  a[p].sum = a[2 * p].sum + a[2 * p + 1].sum;
  a[p].maxv = max(a[2 * p].maxv, a[2 * p + 1].maxv);
}
void change(long long p, long long x, long long d) {
  if (a[p].left == a[p].right) {
    a[p].sum = d;
    a[p].maxv = d;
    return;
  }
  int mid = (a[p].left + a[p].right) / 2;
  if (x <= mid)
    change(2 * p, x, d);
  else if (x > mid)
    change(2 * p + 1, x, d);
  a[p].sum = a[2 * p].sum + a[2 * p + 1].sum;
  a[p].maxv = max(a[2 * p].maxv, a[2 * p + 1].maxv);
}
long long ask(long long p, long long l, long long r) {
  if (a[p].left == l && a[p].right == r) {
    return a[p].sum;
  }
  int mid = (a[p].left + a[p].right) / 2;
  if (r <= mid)
    return ask(2 * p, l, r);
  else if (l > mid)
    return ask(2 * p + 1, l, r);
  else {
    return ask(2 * p, l, mid) + ask(2 * p + 1, mid + 1, r);
  }
}
int main() {
  scanf("%lld%lld", &n, &m);
  for (register int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
  }
  build(1, 1, n);
  for (register int i = 1; i <= m; i++) {
    int x, y, k;
    long long z;
    scanf("%d", &k);
    if (k == 3) {
      scanf("%d%d", &x, &y);
      change(1, x, y);
    } else if (k == 2) {
      scanf("%d%d%lld", &x, &y, &z);
      mod(1, x, y, z);
    } else {
      scanf("%d%d", &x, &y);
      printf("%lld\n", ask(1, x, y));
    }
  }
  return 0;
}