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>
int take() {
  int n;
  scanf("%d", &n);
  return n;
}
double ttake() {
  double n;
  scanf("%lf", &n);
  return n;
}
long long takes() {
  long long n;
  scanf("%lld", &n);
  return n;
}
int cas;
using namespace std;
bool approximatelyEqual(float a, float b, float epsilon) {
  return fabs(a - b) <= ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool essentiallyEqual(float a, float b, float epsilon) {
  return fabs(a - b) <= ((fabs(a) > fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyGreaterThan(float a, float b, float epsilon) {
  return (a - b) > ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyLessThan(float a, float b, float epsilon) {
  return (b - a) > ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
long long tr[4 * 100005], mx[4 * 100005], arr[100005];
void built(int s, int e, int n) {
  if (s > e) return;
  if (s == e) {
    tr[n] = arr[s];
    mx[n] = arr[s];
    return;
  }
  int m = s + (e - s) / 2;
  built(s, m, 2 * n);
  built(m + 1, e, 2 * n + 1);
  tr[n] = tr[2 * n] + tr[2 * n + 1];
  mx[n] = max(mx[2 * n], mx[2 * n + 1]);
}
void upd(int s, int e, int l, int r, int x, int n) {
  if (s > e || e < l || s > r || mx[n] < x) return;
  if (s == e) {
    mx[n] %= x;
    tr[n] %= x;
    return;
  }
  int m = s + (e - s) / 2;
  upd(s, m, l, r, x, 2 * n);
  upd(m + 1, e, l, r, x, 2 * n + 1);
  tr[n] = tr[2 * n] + tr[2 * n + 1];
  mx[n] = max(mx[2 * n], mx[2 * n + 1]);
}
void sets(int s, int e, int l, int r, int x, int n) {
  if (s > e || e < l || s > r) return;
  if (s == e && s == l) {
    mx[n] = x;
    tr[n] = x;
    return;
  }
  int m = s + (e - s) / 2;
  sets(s, m, l, r, x, 2 * n);
  sets(m + 1, e, l, r, x, 2 * n + 1);
  tr[n] = tr[2 * n] + tr[2 * n + 1];
  mx[n] = max(mx[2 * n], mx[2 * n + 1]);
}
long long query(int s, int e, int l, int r, int n) {
  if (s > e || s > r || e < l) return 0;
  if (s >= l && e <= r) return tr[n];
  int m = s + (e - s) / 2;
  return (query(s, m, l, r, 2 * n) + query(m + 1, e, l, r, 2 * n + 1));
}
int main() {
  int tc = 1;
  while (tc--) {
    int n = take(), q = take();
    for (int i = 1; i <= n; i += 1) arr[i] = take();
    built(1, n, 1);
    while (q--) {
      int ty = take();
      if (ty == 1) {
        int l = take(), r = take();
        printf("%lld\n", query(1, n, l, r, 1));
      } else if (ty == 2) {
        int l = take(), r = take(), x = take();
        upd(1, n, l, r, x, 1);
      } else {
        int l = take(), r = l, x = take();
        sets(1, n, l, r, x, 1);
      }
    }
  }
  { return 0; };
}