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;
const long long maxn = 1e5 + 5;
long long n;
long long a[maxn];
struct Tree {
  struct Node {
    long long maxx;
    long long sum;
    inline Node operator=(const long long x) {
      this->maxx = this->sum = x;
      return *this;
    }
  } no[maxn << 2];
  inline void push_up(long long k) {
    no[k].maxx = max(no[(k << 1)].maxx, no[(k << 1 | 1)].maxx);
    no[k].sum = no[(k << 1)].sum + no[(k << 1 | 1)].sum;
  }
  inline void build_tree(long long l, long long r, long long k) {
    if (l == r) {
      no[k] = a[l];
      return;
    }
    long long mid = (l + r) >> 1;
    build_tree(l, mid, (k << 1));
    build_tree(mid + 1, r, (k << 1 | 1));
    push_up(k);
  }
  inline void change(long long l, long long r, long long k, long long K,
                     long long x) {
    if (l == r) {
      no[k] = x;
      return;
    }
    long long mid = (l + r) >> 1;
    if (K <= mid)
      change(l, mid, (k << 1), K, x);
    else
      change(mid + 1, r, (k << 1 | 1), K, x);
    push_up(k);
  }
  inline void change_mod(long long l, long long r, long long k, long long L,
                         long long R, long long x) {
    if (no[k].maxx < x) return;
    if (l == r) {
      no[k] = no[k].sum % x;
      return;
    }
    long long mid = (l + r) >> 1;
    if (L <= mid) change_mod(l, mid, (k << 1), L, R, x);
    if (R > mid) change_mod(mid + 1, r, (k << 1 | 1), L, R, x);
    push_up(k);
  }
  inline long long get_sum(long long l, long long r, long long k, long long L,
                           long long R) {
    if (L <= l && r <= R) return no[k].sum;
    long long ans = 0;
    long long mid = (l + r) >> 1;
    if (L <= mid) ans += get_sum(l, mid, (k << 1), L, R);
    if (R > mid) ans += get_sum(mid + 1, r, (k << 1 | 1), L, R);
    return ans;
  }
} tr;
signed main() {
  long long m;
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  tr.build_tree(1, n, 1);
  long long cmd, l, r, x;
  while (m--) {
    scanf("%lld", &cmd);
    switch (cmd) {
      case 1:
        scanf("%lld%lld", &l, &r);
        printf("%lld\n", tr.get_sum(1, n, 1, l, r));
        break;
      case 2:
        scanf("%lld%lld%lld", &l, &r, &x);
        tr.change_mod(1, n, 1, l, r, x);
        break;
      default:
        scanf("%lld%lld", &l, &x);
        tr.change(1, n, 1, l, x);
    }
  }
  return 0;
}