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 ll = long long;
using namespace std;
const int N = 1e5 + 10;
int n, m, T;
int Mod;
ll Read();
struct node {
  ll val, add, l, r, mul, mx;
} a[N << 2];
ll c[N];
void build(int rt, int l, int r) {
  a[rt].l = l;
  a[rt].r = r;
  a[rt].add = 0;
  a[rt].mul = 1;
  if (l == r) {
    a[rt].mx = a[rt].val = c[l];
    return;
  } else {
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    a[rt].val = a[rt << 1].val + a[rt << 1 | 1].val;
    a[rt].mx = max(a[rt << 1].mx, a[rt << 1 | 1].mx);
  }
}
bool cut(int rt, int x) {
  if (a[rt].mx < x)
    return true;
  else
    return false;
}
void amend(int rt, int l, int r, int opt, ll x) {
  if (opt == 3 && a[rt].l == l && a[rt].r == r) {
    a[rt].val = x;
    a[rt].mx = x;
    return;
  }
  if (l <= a[rt].l && a[rt].r <= r && cut(rt, x) && opt == 2) return;
  if (a[rt].l == a[rt].r && opt == 2) {
    a[rt].val %= x;
    a[rt].mx = a[rt].val;
    return;
  }
  ll mid = (a[rt].l + a[rt].r) >> 1;
  if (l <= mid) amend(rt << 1, l, r, opt, x);
  if (r > mid) amend(rt << 1 | 1, l, r, opt, x);
  a[rt].val = a[rt << 1].val + a[rt << 1 | 1].val;
  a[rt].mx = max(a[rt << 1].mx, a[rt << 1 | 1].mx);
  return;
}
ll query(int rt, int l, int r) {
  if (l <= a[rt].l && a[rt].r <= r) {
    return a[rt].val;
  }
  ll ans = 0, mid = (a[rt].l + a[rt].r) >> 1;
  if (l <= mid) (ans += query(rt << 1, l, r));
  if (r > mid) (ans += query(rt << 1 | 1, l, r));
  a[rt].val = a[rt << 1].val + a[rt << 1 | 1].val;
  a[rt].mx = max(a[rt << 1].mx, a[rt << 1 | 1].mx);
  return ans;
}
int main() {
  n = Read();
  m = Read();
  for (int i = 1; i <= n; i++) c[i] = Read();
  build(1, 1, n);
  for (; m--;) {
    int opt, x, y;
    ll w;
    opt = Read();
    if (opt == 1) {
      x = Read(), y = Read();
      printf("%lld\n", query(1, x, y));
    } else if (opt == 2) {
      x = Read(), y = Read(), w = Read();
      amend(1, x, y, opt, w);
    } else if (opt == 3) {
      x = Read(), w = Read();
      amend(1, x, x, opt, w);
    }
  }
  return 0;
}
ll Read() {
  ll val = 0, opt = 1;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') opt = -1;
  while (isdigit(ch)) (val *= 10) += ch - '0', ch = getchar();
  return val * opt;
}