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 <typename tp>
inline void read(tp &x) {
  x = 0;
  register int f = 1;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  x *= f;
}
template <typename tp>
inline void print(tp x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 1e6 + 10;
long long n, m;
long long opt, in_x, in_y;
long long a[N];
struct node {
  long long sum;
  long long maxx;
};
struct seg_tree {
  node tr[N << 1];
  inline void pushup(long long rt) {
    tr[rt].maxx = max(tr[rt << 1].maxx, tr[rt << 1 | 1].maxx);
    tr[rt].sum = tr[rt << 1].sum + tr[rt << 1 | 1].sum;
  }
  inline void build(long long l, long long r, long long rt) {
    if (l == r) {
      tr[rt].sum = tr[rt].maxx = a[l];
      return;
    }
    long long mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  inline void update(long long pos, long long c, long long l, long long r,
                     long long rt) {
    if (l == r) {
      tr[rt].maxx = tr[rt].sum = c;
      return;
    }
    long long mid = (l + r) >> 1;
    if (pos <= mid)
      update(pos, c, l, mid, rt << 1);
    else
      update(pos, c, mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  inline void update_mod(long long L, long long R, long long c, long long l,
                         long long r, long long rt) {
    if (l == r) {
      tr[rt].maxx %= c;
      tr[rt].sum %= c;
      return;
    }
    long long mid = (l + r) >> 1;
    if (L <= mid && tr[rt << 1].maxx >= c) update_mod(L, R, c, l, mid, rt << 1);
    if (R > mid && tr[rt << 1 | 1].maxx >= c)
      update_mod(L, R, c, mid + 1, r, rt << 1 | 1);
    pushup(rt);
  }
  inline long long query(long long L, long long R, long long l, long long r,
                         long long rt) {
    if (L <= l && r <= R) return tr[rt].sum;
    long long mid = (l + r) >> 1;
    long long ret = 0;
    if (L <= mid) ret += query(L, R, l, mid, rt << 1);
    if (R > mid) ret += query(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
  }
} SEG;
signed main() {
  read(n), read(m);
  for (register long long i = 1; i <= n; ++i) read(a[i]);
  SEG.build(1, n, 1);
  while (m--) {
    read(opt);
    if (opt == 1) {
      read(in_x), read(in_y);
      print(SEG.query(in_x, in_y, 1, n, 1)), putchar('\n');
    } else if (opt == 2) {
      long long c;
      read(in_x), read(in_y), read(c);
      SEG.update_mod(in_x, in_y, c, 1, n, 1);
    } else {
      read(in_x), read(in_y);
      SEG.update(in_x, in_y, 1, n, 1);
    }
  }
  return 0;
}