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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
long long a, b, c, d[100007], n, m, T[100007 * 4], M[100007 * 4], type;
void form_tree(int l, int r, int v) {
  if (l == r) {
    T[v] = M[v] = d[l];
    return;
  }
  form_tree(l, (l + r) / 2, v * 2);
  form_tree((l + r) / 2 + 1, r, v * 2 + 1);
  T[v] = T[v * 2] + T[v * 2 + 1];
  M[v] = max(M[v * 2], M[v * 2 + 1]);
}
long long bul(int x, int y, int l, int r, int v) {
  if (y < l || r < x) return 0;
  if (x <= l && r <= y) return T[v];
  return bul(x, y, l, (l + r) / 2, v * 2) +
         bul(x, y, (l + r) / 2 + 1, r, v * 2 + 1);
}
void upd(int x, int y, long long z, int l, int r, int v) {
  if (M[v] < z) return;
  if (y < l || r < x) return;
  if (l == r) {
    T[v] %= z;
    M[v] %= z;
    return;
  }
  upd(x, y, z, l, (l + r) / 2, v * 2);
  upd(x, y, z, (l + r) / 2 + 1, r, v * 2 + 1);
  T[v] = T[v * 2] + T[v * 2 + 1];
  M[v] = max(M[v * 2], M[v * 2 + 1]);
}
void upd1(int x, long long z, int l, int r, int v) {
  if (l == r) {
    T[v] = M[v] = z;
    return;
  }
  if (x <= (l + r) / 2)
    upd1(x, z, l, (l + r) / 2, v * 2);
  else
    upd1(x, z, (l + r) / 2 + 1, r, v * 2 + 1);
  T[v] = T[v * 2] + T[v * 2 + 1];
  M[v] = max(M[v * 2], M[v * 2 + 1]);
}
int main() {
  scanf("%lld%lld", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", d + i);
  form_tree(1, n, 1);
  while (m--) {
    scanf("%lld", &type);
    if (type == 1) {
      scanf("%lld%lld", &a, &b);
      printf("%lld\n", bul(a, b, 1, n, 1));
    } else if (type == 2) {
      scanf("%lld%lld%lld", &a, &b, &c);
      upd(a, b, c, 1, n, 1);
    } else {
      scanf("%lld%lld", &a, &c);
      upd1(a, c, 1, n, 1);
    }
  }
  return 0;
}