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;
void tempwj() {
  freopen("P3380_2.in", "r", stdin);
  freopen("hash.out", "w", stdout);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long qpow(long long a, long long b, long long mod) {
  a %= mod;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans % mod;
}
struct cmp {
  bool operator()(const pair<int, long long>& a,
                  const pair<int, long long>& b) {
    return a.second > b.second;
  }
};
int lb(int x) { return x & -x; }
const long long inf = 0x3f3f3f;
const double esp = 1e-9;
const long long INF = 0x3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 10;
struct Node {
  int l, r, num;
  long long sum;
} node[maxn << 2];
int a[maxn];
void build(int l, int r, int no) {
  node[no].l = l;
  node[no].r = r;
  if (l == r) {
    node[no].num = a[l];
    node[no].sum = a[l];
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, no << 1);
  build(mid + 1, r, no << 1 | 1);
  node[no].num = max(node[no << 1].num, node[no << 1 | 1].num);
  node[no].sum = node[no << 1].sum + node[no << 1 | 1].sum;
}
void update(int l, int r, int no, int num) {
  if (node[no].l > r || node[no].r < l) return;
  if (node[no].num < num) return;
  if (node[no].l == node[no].r) {
    node[no].num %= num;
    node[no].sum %= num;
    return;
  }
  update(l, r, no << 1, num);
  update(l, r, no << 1 | 1, num);
  node[no].num = max(node[no << 1].num, node[no << 1 | 1].num);
  node[no].sum = node[no << 1].sum + node[no << 1 | 1].sum;
}
void update2(int pos, int no, int num) {
  if (node[no].l > pos || node[no].r < pos) return;
  if (node[no].l == node[no].r) {
    node[no].num = num;
    node[no].sum = num;
    return;
  }
  update2(pos, no << 1, num);
  update2(pos, no << 1 | 1, num);
  node[no].num = max(node[no << 1].num, node[no << 1 | 1].num);
  node[no].sum = node[no << 1].sum + node[no << 1 | 1].sum;
}
long long query(int l, int r, int no) {
  if (node[no].l > r || node[no].r < l) return 0;
  if (node[no].l >= l && node[no].r <= r) return node[no].sum;
  return query(l, r, no << 1) + query(l, r, no << 1 | 1);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, n, 1);
  while (m--) {
    int f, l, r;
    scanf("%d%d%d", &f, &l, &r);
    if (f == 1) {
      long long ans = query(l, r, 1);
      printf("%lld\n", ans);
    } else if (f == 2) {
      int x;
      scanf("%d", &x);
      update(l, r, 1, x);
    } else {
      update2(l, 1, r);
    }
  }
}