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;
int n, n1;
struct node {
  long long s;
  int m;
  int j;
};
node tree[400000];
void update(int node) {
  if (!node) return;
  int i = node;
  tree[i].s = tree[i * 2].s + tree[i * 2 + 1].s;
  if (tree[i * 2].m >= tree[i * 2 + 1].m) {
    tree[i].m = tree[i * 2].m;
    tree[i].j = tree[i * 2].j;
  } else {
    tree[i].m = tree[i * 2 + 1].m;
    tree[i].j = tree[i * 2 + 1].j;
  }
  update(node / 2);
}
int maximum(int l, int r, int left, int right, int node) {
  if (l == left && r == right) return tree[node].j;
  int m = (left + right) / 2;
  if (r <= m) {
    return maximum(l, r, left, m, node * 2);
  }
  if (m < l) return maximum(l, r, m + 1, right, node * 2 + 1);
  int m1 = maximum(l, m, left, m, node * 2),
      m2 = maximum(m + 1, r, m + 1, right, node * 2 + 1);
  if (tree[n1 + m1 - 1].m > tree[n1 + m2 - 1].m) return m1;
  return m2;
}
long long sum(int l, int r, int left, int right, int node) {
  if (l == left && r == right) return tree[node].s;
  int m = (left + right) / 2;
  if (r <= m) {
    return sum(l, r, left, m, node * 2);
  }
  if (m < l) return sum(l, r, m + 1, right, node * 2 + 1);
  return sum(l, m, left, m, node * 2) +
         sum(m + 1, r, m + 1, right, node * 2 + 1);
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  int i, J;
  n1 = 1;
  while (n1 < n) n1 *= 2;
  for (i = 0; i < n; ++i) {
    scanf("%lld", &J);
    tree[n1 + i].s = 1ll * J;
    tree[n1 + i].m = J;
    tree[n1 + i].j = 1 + i;
  }
  for (i = n1 - 1; i > 0; --i) {
    tree[i].s = tree[i * 2].s + tree[i * 2 + 1].s;
    if (tree[i * 2].m >= tree[i * 2 + 1].m) {
      tree[i].m = tree[i * 2].m;
      tree[i].j = tree[i * 2].j;
    } else {
      tree[i].m = tree[i * 2 + 1].m;
      tree[i].j = tree[i * 2 + 1].j;
    }
  }
  int t, a, b, c, x;
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &a, &b);
      printf("%lld\n", sum(a, b, 1, n1, 1));
    }
    if (t == 2) {
      scanf("%d%d%d", &a, &b, &c);
      x = maximum(a, b, 1, n1, 1);
      while (tree[n1 + x - 1].m >= c) {
        tree[n1 + x - 1].m %= c;
        tree[n1 + x - 1].s %= c;
        update((n1 + x - 1) / 2);
        x = maximum(a, b, 1, n1, 1);
      }
    }
    if (t == 3) {
      scanf("%d%d", &a, &b);
      tree[n1 + a - 1].m = b;
      tree[n1 + a - 1].s = 1ll * b;
      update((n1 + a - 1) / 2);
    }
  }
  return 0;
}