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;
struct Node {
  long long maxx, sum;
} node[100010 << 2];
void pushup(int rt) {
  node[rt].maxx = max(node[rt << 1].maxx, node[rt << 1 | 1].maxx);
  node[rt].sum = node[rt << 1].sum + node[rt << 1 | 1].sum;
}
void build(int l, int r, int rt) {
  if (l == r) {
    scanf("%lld", &node[rt].maxx);
    node[rt].sum = node[rt].maxx;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void point_update(int index, int k, int l, int r, int rt) {
  if (l == r) {
    node[rt].maxx = k;
    node[rt].sum = k;
    return;
  }
  int m = (l + r) >> 1;
  if (index <= m)
    point_update(index, k, l, m, rt << 1);
  else
    point_update(index, k, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void interval_update(int L, int R, int mod, int l, int r, int rt) {
  if (L > r || R < l) return;
  if (node[rt].maxx < mod) return;
  if (l == r) {
    node[rt].sum %= mod;
    node[rt].maxx = node[rt].sum;
    return;
  }
  int m = (l + r) >> 1;
  interval_update(L, R, mod, l, m, rt << 1);
  interval_update(L, R, mod, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (L > r || R < l) return 0;
  if (L <= l && r <= R) return node[rt].sum;
  long long sum = 0;
  int m = (l + r) >> 1;
  sum += query(L, R, l, m, rt << 1);
  sum += query(L, R, m + 1, r, rt << 1 | 1);
  return sum;
}
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    int opt, l, r, mod;
    scanf("%d", &opt);
    if (opt == 3) {
      scanf("%d%d", &l, &r);
      point_update(l, r, 1, n, 1);
    }
    if (opt == 2) {
      scanf("%d%d%d", &l, &r, &mod);
      interval_update(l, r, mod, 1, n, 1);
    }
    if (opt == 1) {
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(l, r, 1, n, 1));
    }
  }
  return 0;
}