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;
long long int tree_max[4 * 100009], tree_sum[4 * 100009];
int a[100009];
void build_sum(int node, int start, int end) {
  if (start == end) {
    tree_sum[node] = a[start];
    return;
  }
  int mid = (start + end) >> 1;
  build_sum(2 * node + 1, start, mid);
  build_sum(2 * node + 2, mid + 1, end);
  tree_sum[node] = tree_sum[2 * node + 1] + tree_sum[2 * node + 2];
}
void build_max(int node, int start, int end) {
  if (start == end) {
    tree_max[node] = a[start];
    return;
  }
  int mid = (start + end) >> 1;
  build_max(2 * node + 1, start, mid);
  build_max(2 * node + 2, mid + 1, end);
  tree_max[node] = max(tree_max[2 * node + 1], tree_max[2 * node + 2]);
}
void point_update(int node, int start, int end, int idx, int val) {
  if (start == end) {
    tree_sum[node] = val;
    tree_max[node] = val;
    a[start] = val;
    return;
  }
  int mid = (start + end) >> 1;
  if (idx > mid) {
    point_update(2 * node + 2, mid + 1, end, idx, val);
  } else {
    point_update(2 * node + 1, start, mid, idx, val);
  }
  tree_sum[node] = tree_sum[2 * node + 1] + tree_sum[2 * node + 2];
  tree_max[node] = max(tree_max[2 * node + 1], tree_max[2 * node + 2]);
}
long long int sum_query(int node, int start, int end, int l, int r) {
  if (l > end || r < start) {
    return 0;
  }
  if (start >= l && end <= r) {
    return tree_sum[node];
  }
  int mid = (start + end) >> 1;
  long long int p1 = sum_query(2 * node + 1, start, mid, l, r);
  long long int p2 = sum_query(2 * node + 2, mid + 1, end, l, r);
  return p1 + p2;
}
void mod_update(int node, int start, int end, int l, int r, int val) {
  if (l > end || r < start) {
    return;
  }
  if (start == end) {
    tree_sum[node] %= val;
    tree_max[node] %= val;
    a[start] %= val;
    return;
  }
  if (l > end || r < start) {
    return;
  }
  int mid = (start + end) >> 1;
  if (tree_max[2 * node + 1] >= val) {
    mod_update(2 * node + 1, start, mid, l, r, val);
  }
  if (tree_max[2 * node + 2] >= val) {
    mod_update(2 * node + 2, mid + 1, end, l, r, val);
  }
  tree_sum[node] = tree_sum[2 * node + 1] + tree_sum[2 * node + 2];
  tree_max[node] = max(tree_max[2 * node + 1], tree_max[2 * node + 2]);
}
int main() {
  int i, j, n, m, k, x, y, z;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  build_max(0, 0, n - 1);
  build_sum(0, 0, n - 1);
  while (m--) {
    scanf("%d", &x);
    if (x == 1) {
      scanf("%d%d", &y, &z);
      y--;
      z--;
      long long int zz = sum_query(0, 0, n - 1, y, z);
      printf("%lld\n", zz);
      ;
    } else if (x == 2) {
      scanf("%d%d%d", &x, &y, &z);
      x--;
      y--;
      mod_update(0, 0, n - 1, x, y, z);
    } else {
      scanf("%d%d", &k, &y);
      k--;
      point_update(0, 0, n - 1, k, y);
    }
  }
}