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;
inline long long read() {
  long long f = 1, sum = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return f * sum;
}
inline void print_num(long long x) {
  if (x > 9) print_num(x / 10);
  putchar(x % 10 + '0');
}
inline void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  print_num(x);
  puts("");
}
long long tree[400010], tree1[400010], a[100010], n, m;
inline long long ls(long long x) { return x << 1; }
inline long long rs(long long x) { return x << 1 | 1; }
inline void pushup(long long x) {
  tree[x] = tree[ls(x)] + tree[rs(x)];
  tree1[x] = max(tree1[ls(x)], tree1[rs(x)]);
}
inline void build(long long l, long long r, long long num) {
  if (l == r) {
    tree[num] = tree1[num] = a[l];
    return;
  }
  build(l, ((l + r) >> 1), ls(num)), build(((l + r) >> 1) + 1, r, rs(num));
  pushup(num);
}
inline void update(long long l, long long r, long long x, long long y,
                   long long k, long long num) {
  if (tree1[num] < k) return;
  if (l == r && l >= x && r <= y) {
    tree[num] %= k, tree1[num] = tree[num];
    return;
  }
  if (((l + r) >> 1) >= x) update(l, ((l + r) >> 1), x, y, k, ls(num));
  if (((l + r) >> 1) < y) update(((l + r) >> 1) + 1, r, x, y, k, rs(num));
  pushup(num);
}
inline void update1(long long l, long long r, long long x, long long k,
                    long long num) {
  if (l == r && l == x) {
    tree[num] = tree1[num] = k;
    return;
  }
  if (((l + r) >> 1) >= x)
    update1(l, ((l + r) >> 1), x, k, ls(num));
  else
    update1(((l + r) >> 1) + 1, r, x, k, rs(num));
  pushup(num);
}
inline long long getsum(long long l, long long r, long long x, long long y,
                        long long num) {
  if (l >= x && r <= y) {
    return tree[num];
  }
  long long sum = 0;
  if (((l + r) >> 1) >= x) sum += getsum(l, ((l + r) >> 1), x, y, ls(num));
  if (((l + r) >> 1) < y) sum += getsum(((l + r) >> 1) + 1, r, x, y, rs(num));
  return sum;
}
signed main() {
  n = read(), m = read();
  for (register long long i = 1; i <= n; i = -~i) a[i] = read();
  build(1, n, 1);
  for (register long long i = 1; i <= m; i = -~i) {
    long long opo = read();
    if (opo == 1) {
      long long x = read(), y = read();
      print(getsum(1, n, x, y, 1));
    } else if (opo == 2) {
      long long x = read(), y = read(), p = read();
      update(1, n, x, y, p, 1);
    } else {
      long long x = read(), k = read();
      update1(1, n, x, k, 1);
    }
  }
  return 0;
}