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 int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^= 48), ch = getchar();
  return f ? -x : x;
}
int n, m;
struct node {
  int L, R, maxn;
  long long sum;
  node *lc, *rc;
};
void build(node *now, int l, int r) {
  now->L = l;
  now->R = r;
  now->sum = 0;
  now->maxn = -21474836;
  if (l + 1 < r) {
    now->lc = new node;
    now->rc = new node;
    int mid = (l + r) >> 1;
    build(now->lc, l, mid);
    build(now->rc, mid, r);
  } else
    now->lc = now->rc = NULL;
}
void add(node *now, int x, int k) {
  if (now->L + 1 == now->R) {
    now->sum = k;
    now->maxn = k;
  } else {
    int mid = (now->L + now->R) >> 1;
    if (x < mid) add(now->lc, x, k);
    if (x >= mid) add(now->rc, x, k);
    now->sum = now->lc->sum + now->rc->sum;
    now->maxn = max(now->lc->maxn, now->rc->maxn);
  }
}
bool f = 1;
int check_max(node *now, int l, int r) {
  if (l <= now->L && now->R <= r + 1)
    return now->maxn;
  else {
    int mid = (now->L + now->R) >> 1;
    int ans = -21474836;
    if (l < mid) ans = max(ans, check_max(now->lc, l, r));
    if (r >= mid) ans = max(ans, check_max(now->rc, l, r));
    return ans;
  }
}
void modd(node *now, int l, int r, int mod) {
  if (l <= now->L && now->R <= r + 1 && (check_max(now, l, r) < mod)) return;
  if (l <= now->L && now->R <= r + 1 && now->L + 1 == now->R) {
    now->sum %= mod;
    now->maxn %= mod;
    return;
  }
  int mid = (now->L + now->R) >> 1;
  if (l < mid) modd(now->lc, l, r, mod);
  if (r >= mid) modd(now->rc, l, r, mod);
  now->sum = now->lc->sum + now->rc->sum;
  now->maxn = max(now->lc->maxn, now->rc->maxn);
}
long long check_sum(node *now, int l, int r) {
  if (l <= now->L && now->R <= r + 1)
    return now->sum;
  else {
    int mid = (now->L + now->R) / 2;
    long long ans = 0;
    if (l < mid) ans += (long long)check_sum(now->lc, l, r);
    if (r >= mid) ans += (long long)check_sum(now->rc, l, r);
    return ans;
  }
}
int main() {
  n = read();
  m = read();
  node *root;
  root = new node;
  build(root, 1, n + 1);
  for (int i = 1; i <= n; ++i) {
    int x = read();
    add(root, i, x);
  }
  for (int i = 1; i <= m; ++i) {
    int opt = read();
    if (opt == 1) {
      int l = read(), r = read();
      long long ans = check_sum(root, l, r);
      printf("%lld\n", ans);
    } else if (opt == 2) {
      int l = read(), r = read(), mod = read();
      modd(root, l, r, mod);
    } else if (opt == 3) {
      int x = read(), k = read();
      add(root, x, k);
    }
  }
  return 0;
}