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 a[1000005];
struct seg {
  long long sum;
  long long mmax;
} tree[1000005];
void makeTree(int node, int b, int e) {
  int m = (b + e) / 2, left = node << 1;
  int right = left + 1;
  if (b == e) {
    tree[node].sum = a[b];
    tree[node].mmax = a[b];
    return;
  }
  makeTree(left, b, m);
  makeTree(right, m + 1, e);
  tree[node].sum = tree[left].sum + tree[right].sum;
  tree[node].mmax = max(tree[left].mmax, tree[right].mmax);
}
void update1(int node, int b, int e, int indx, int val) {
  int m = (b + e) / 2, left = node << 1;
  int right = left + 1;
  if (b > indx || e < indx) return;
  if (b == e) {
    tree[node].sum = val;
    tree[node].mmax = val;
    return;
  }
  update1(left, b, m, indx, val);
  update1(right, m + 1, e, indx, val);
  tree[node].sum = tree[left].sum + tree[right].sum;
  tree[node].mmax = max(tree[left].mmax, tree[right].mmax);
}
void update2(int node, int b, int e, int l, int r, int x) {
  int m = (b + e) / 2, left = node << 1;
  int right = left + 1;
  if (b > r || e < l) return;
  if (tree[node].mmax < x) return;
  if (b == e) {
    tree[node].sum %= x;
    tree[node].mmax %= x;
    return;
  }
  update2(left, b, m, l, r, x);
  update2(right, m + 1, e, l, r, x);
  tree[node].sum = tree[left].sum + tree[right].sum;
  tree[node].mmax = max(tree[left].mmax, tree[right].mmax);
}
long long query(int node, int b, int e, int l, int r) {
  int m = (b + e) / 2, left = node << 1;
  int right = left + 1;
  if (b > r || e < l) return 0;
  if (b >= l && e <= r) return tree[node].sum;
  return query(left, b, m, l, r) + query(right, m + 1, e, l, r);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  makeTree(1, 1, n);
  int l, r, mod, val;
  while (q--) {
    int flag;
    scanf("%d", &flag);
    if (flag == 1) {
      scanf("%d%d", &l, &r);
      long long res = query(1, 1, n, l, r);
      printf("%lld\n", res);
      continue;
    }
    if (flag == 2) {
      scanf("%d%d%d", &l, &r, &mod);
      update2(1, 1, n, l, r, mod);
      continue;
    }
    scanf("%d%d", &l, &val);
    update1(1, 1, n, l, val);
  }
}